ELEMENTS  OF  ELECTRICITY 


FOR 


TECHNICAL  STUDENTS 


BY 

W.   H.    TIMBIE 

INSTRUCTOR    IN    APPLIED    SCIENCE,    WENTWORTH    INSTITUTE,    BOSTON,    MASS. 
FORMERLY   INSTRUCTOR  IN    INDUSTRIAL    ELECTRICITY,    PRATT    INSTITUTE 


FIRST    EDITION 

SECOND    THOUSAND 


NEW   YORK 

JOHN  WILEY   &   SONS     . 
LONDON:    CHAPMAN  &  HALL,   LIMITED 
1911 


Copyright,  1910 

BT 

W.   H.   TIMBIE 


THE    SCIENTIFIC    PRESS 

ROBERT   ORUMMONO    AND    COMPANY 

BROOKLYN,   N.   V. 


PREFACE 


THE  following  text  is  designed  primarily  to  meet  the  needs 
of  young  men  who  desire  to  follow  an  occupation  connected 
with  the  electrical,  or  allied,  industries.  Such  young  men 
need  a  thorough  grounding  in  the  fundamental  principles 
of  electrical  theory  and  measurement.  They  usually  have 
neither  the  time  nor  the  previous  preparation  for  acquiring 
this  knowledge  through  advanced  or  mathematical  treatises. 
They  need  their  information  in  a  directly  usable  form. 
After  acquiring  the  words  of  a  law,  they  need,  above  all, 
to  know  what  the  law  means  in  a  practical  sense.  Thus 
they  need  extended  drill  in  applying  the  same  underlying 
principle  to  the  many  diversified  problems  of  the  electrical 
industries. 

This  text  is  based  fundamentally  upon  a  series  of  mimeo- 
graphed notes  which  have  been  used  for  several  years  with 
both  day  and  evening  classes  in  Applied  and  Industrial 
Electricity  at  Pratt  Institute.  It  may  be  justly  claimed 
by  the  author,  therefore,  that  his  presentation,  if  not  the 
usual  one,  or  to  some  minds,  in  logical  order,  is  at  least 
teachable;  that  it  secures  results  which  are  direct  and  which 
constitute  a  desirable  preparation  for  more  advanced  study, 
as  well  as  for  practical  work  in  the  many  lines  of  the  elec- 
trical world. 

In  presenting  each  idea,  it  will  be  noted,  such  language 
is  used  as  will  make  maximum  use  of  the  student's  present 
knowledge  or  observation;  a  brief  definition  or  discussion 
being  followed  immediately  by  a  number  of  simple  problems 
sufficient  not  only  to  enable  the  student  to  make  the  "  sense 
of  the  matter  "  his  own,  in  his  own  language  and  manner  of 
thought,  but  also  to  cultivate  in  him  some  degree  of  alert- 

224449  m 


iv  PREFACE 

ness  and  initiative  in  applying  his  knowledge.  In  addi- 
tion, at  the  close  of  each  chapter,  a  number  of  more 
advanced  problems  is  given  which  have  been  in  the  process 
of  collection  and  modification  for  a  number  of  years. 

The  order  followed,  as  has  been  suggested,  is  merely  that 
which  the  author  has  found  most  practicable,  and  which 
will — by  means  of  the  amount  of  general  information  as 
to  instruments,  etc.,  placed  early  in  the  book — render  the 
parallel  laboratory  course  most  efficient  as  a  medium  for 
teaching,  illustration,  and  independent  growth  on  the 
part  of  the  student. 

The  author  believes  that  the  book  has  distinctively 
these  four  desirable  qualities: 

(1)  It   contains   an   adequate   amount   of  information 
concerning  electrical  laws  and  practice  in  an  immediately 
usable  form. 

(2)  It  applies  the  information  contained  to  real  things 
and  not  to  abstract  theory  and  conditions. 

(3)  It    provides  sufficient   drill  in    concrete    practical 
problems,  which  not  only  afford  in  themselves  an  addi- 
tional amount  of  profitable  information,  but  also  develop 
in  the  student  a  capacity  for  applying  what  he  has  learned ; 
the  large  number  of  diagrams  aiding  materially  in  pro- 
viding exact  and  clean-cut  conceptions  of  the  conditions 
of  the  problems. 

(4)  It  presents,   out  of  the  great  mass  of  electrical 
phenomena,    only   those   facts    and   principles   which    a 
technica1    student   needs   to   know,    and   to   know   well. 
History  and  general  theory,  interesting  as  they  are,  have 
been  omitted. 

The  text  is  confined  to  what  is  believed  to  be  an  adequate 
treatment  of  a  few  fundamental  ideas  rather  than  to  a 
discursive  treatment  of  many.  Therefore  it  includes 
only  ideas  of  the  elementary  principles  of  Direct  and  Alter- 
nating Current  generation,  distribution  and  utilization 
in  light  and  power.  This  affords  a  sure  foundation  on  which 


PREFACE  v 

to  build  more  advanced  courses,  and  yet  presents  a  practical 
and  well-rounded  course  for  the  man  who  is  to  receive  no 
further  electrical  education. 

It  has  been  found  that  about  fifteen  weeks  are  needed 
for  the  student  to  satisfactorily  cover  the  subject  matter 
of  the  text,  allotting  five  hours  per  week  to  recitations  and 
six  hours  to  laboratory  work.  But  the  book  is  also  arranged 
so  that  by  omitting  the  matter  in  smaller  type  and  in  Chap- 
ters VI,  IX,  and  XI,  a  consecutive  course  can  be  given 
in  twelve  weeks,  consisting  of  only  three  hours  recitation 
and  four  hours  laboratory  work. 

The  previous  preparation  required  is  a  knowledge  of 
common  school  arithmetic  and  of  the  simplest  algebra. 
Alternating  current  is  presented  without  the  use  of  even 
trigonometry,  although  the  equations  are  given  in  each 
case  for  trignometric  solution  of  problems.  All  the  trig- 
ometry  needed  is  explained  on  two  pages  of  the  Appendix. 
The  elementary  facts  of  Mechanics  and  of  Heat  may  be 
taken  up  before  studying  the  text,  or  may  be  supplied  by  the 
instructor  from  time  to  time  as  there  is  need. 

The  book  then  is  especially  adapted  for  use  in  the  short, 
practical  courses  given  in  Trade,  Industrial,  and  Technical 
High  Schools  and  Apprenticeship  courses,  but  also  affords 
a  substantial  ground-work  for  the  more  advanced  work 
in  Electrical  theory  and  practice  in  colleges  and  universities. 

In  conclusion,  the  author  wishes  to  express  his  appreciation 
and  thanks  to  Mr.  Joseph  M.  Jameson,  Head  of  the  Depart- 
ment of  Physics,  and  to  Mr.  Arthur  L.  Williston,  formerly 
Director  of  the  School  of  Science  and  Technology,  Pratt 
Institute,  for  their  assistance  and  constant  encouragement 
in  developing  an  Elementary  Electrical  course  as  here 
outlined.  Grateful  acknowledgment-  is  also  extended  to 
my  colleagues,  Mr.  John  A.  Randall  and  Mr.  Warren  L. 
Harlow  for  many  valuable  criticisms  and  suggestions. 

W.  H.  TlMBIE. 
BROOKLYN,  N.  Y.,  November,  1910. 


ACKNOWLEDGMENTS 

Pages  526-540,  Appendix,  J.   M.  Jameson's  "  Elementary 
Practical  Mechanics." 

Page  544,  Table  of  Equivalent  Values,   Kent's  "  Mechan- 
ical Engineers'  Pocketbook." 

Page  541,  Table  of    RH  Values,   Caldwell's  "Electrical 
Problems." 

Page  332,  Table  of  Electrochemical  Equivalents,  "  Stand- 
ard Handbook  for  Electrical  Engineers." 

Page     152,    Table     of    Hysteresis    Constants,     Foster's 
"  Electrical  Engineer's  Pocketbook." 

Figs.   204,   229,   230,   304,   304a,  Prof.    V.     Karapetoff's 
-•'Experimental  Electrical  Engineering." 

Figs.  39,  40,  265,  266,  385a,    396a,  3966,    411,   General 
Electric  Co. 

Figs.  95,  140,  149,  150,  167,  174,  179,  239,  Westinghouse 
Electrical  and  Mfg.  Co. 

Figs.  34  and  35,  Browning  Engineering  Co. 

Figs.  114,   115,   116,   117,   118,   119,  120,  231,  232,  287, 
Leeds  &  Northrup  Co. 

Figs.  141,  Allis  Chalmers  Co. 

Fig.  170,  Stow  Mfg.  Co. 

Figs.  172,  172a,  Cutler  Hammer  Co. 

Fig.  173a,  Ward,  Leonard  Co. 

Figs.  205,  206,  207,  Fort  Wayne  Electric  Works. 

Fig.  271,  Queen  &  Co. 

Figs.  279,  280,  Bristol  Co. 

Figs.  284,  291,  294,  301,  Weston  Elec.  Instrument  Co. 

Fig.  289,  Switchboard  Equipment  Co. 

Fig.  355a,  Moore  Light  Co. 

vi 


TABLE  OF  CONTENTS 


CHAPTER   I 
MAGNETS  AND  MAGNETISM 

PAGE 

Definition  of  Magnet  —  Poles  —  Magnetic  Lines  —  Flux  —  Field 
Intensity — Nature  of  Magnetic  Lines — Attraction  and  Repul- 
sion of  Magnets — Permeability — Induction — Magnetic  Screens 
— Compass — Unit  Poles — Measure  of  Mutual  Action  between 
Two  Magnets — Mathematical  Relation  between  Unit  Pole  and 
Force  Line — Two  Equations  for  Magnetic  Force — Ring 
Magnets — Consequent  Poles — Permanent  Magnets — Magnetic 
Molecule  Theory. 1 

CHAPTER   II 

ELECTROMAGNETS 

Magnetic  Field  about  a  Straight  Wire — Thumb  Rule — Resultant 
of  Circular  and  Parallel  Fields — Moving  Force  in  Voltmeter, 
Ammeter,  Motors,  etc. — Field  about  a  Coil — Thumb  Rule  for 
Coil — Electromagnets — Magnetic  Hoists — Telegraph — Gener- 
ator and  Motor  Fields — Sucking  Coils — Non-inductive  Coils.  .  24 

CHAPTER   III 
OHM'S  LAW 

The  Electric  Current — Units;  Ampere  (current);  Volt  (pressure); 
Ohm  (resistance) — Analogy  of  Electrical  to  Hydraulic  Units 
— Diagrams  of  Electric  Circuits — Table  of  Convenient 
Symbols— Voltage,  the  Essential  Factor— Potential— Electri- 
cal Potential — Brush  Potential —Fall  of  Potential  along  a 
Uniform  Wire — Ohm's  Law — Applications  of  Ohm's  Law — 

vii 


viii  TABLE  OF  CONTENTS 


PAGE 

Series  and  Parallel  Circuits — Voltage,  Current,  and  Resistance 
Relations  in  Series  and  Parallel  Circuits — Conductance  ....         37 


CHAPTER   IV 
POWER  MEASUREMENT 

Use  of  Ammeter,  Voltmeter,  and  Wattmeter — Electric  Power; 
Watt;  Kilowatt — Variations  of  Power  Equation — Electric 
Energy;  Kilowatt-hour;  Watt-second  or  Joule — Electric 
Energy  Converted  to  Heat  Energy — Heat  Equivalent  of 
Electricity — Efficiency:  of  Electric  Machines;  of  Electric 
Transmission ;  of  Electric  Lamps 70 

CHAPTER   V 

MEASUREMENT  OF  RESISTANCE 

Circular  Measure — Mil — Mil-foot — Resistivity — Temperature  Co- 
efficient of  Resistance — Use  of  Wire  Tables — Temperature 
Measured  by  Change  in  Resistance — Methods  of  Measuring 
Resistance;  Fall  of  Potential,  Ammeter-voltmeter,  Wheat- 
stone  Bridge,  Voltmeter  Method — Insulation  Resistance  by 
Galvanometer  Deflection  and  Standard  Megohm — Location 
of  Faults 95 

CHAPTER  VI 
MAGNETIC  FIELD  DUE  TO  A  CURRENT 

Field  within  a  Coil — Am pe re-turns — Magnetomotive  Force.  Re- 
luctance— Ohm's  Law  of  the  Magnetic  Current — Relation 
between  B  and  H — Permeability  at  Different  Degrees  of 
Magnetization — Three  Stages  of  Magnetization — Saturation 
Point— Hysteresis 130 

CHAPTER  VII 
THE  GENERATOR 

Electromagnetic  Induction — Direction  of  Induced  E.M.F. — 
Amount  of  Induced  E.M.F. — Current  in  Revolving  Loop — 
Sine  Curve  of  E.M.F,— Collecting  Rings— A .C.  Power — Com- 


TABLE  OF  CONTENTS  ix 

PAGE 

mutator — D.C.  Power — Ring  and  Drum  Armatures — Action 
within  an  Armature — Magnetization  of  the  Core — Neutral 
Axis — Axis  of  Least  Sparking — Voltage  and  Resistance  of 
D.C.  Armature — Field  Excitation  of  Generator — Separately 
Excited,  Series  Wound,  Shunt  Wound,  Compound — Losses  in 
Generator — Eddy  Current  Loss 159 

CHAPTER   VIII 
THE  MOTOR 

Force  "on  Wire  in  Magnetic  Field — Torque  of  Motor — Power 
Necessary  to  Drive  Generators — Damping  of  Electrical 
Instruments — Counter  E.M.F. — Armature  Reaction  in  Motors 
— Direction  of  Rotation  of  Motor — Shunt  Motor — Starting 
Box — Speed  Regulation  and  Control  of  Shunt  Motor — No- 
Voltage  Release — Overload  Release — Starting  and  Stopping 
Shunt  Motor — Series  Motor:  Starting — Series- Parallel  Con- 
trol— Compound:  Differential  and  Cumulative — Motor  and 
Generator  Characteristics  Compared 199 

CHAPTER  IX 
FURTHER  APPLICATIONS 

SOLUTION   OF   SOME   OF   THE    MORE    DIFFICULT    PROBLEMS   ENCOUNTERED 
IN   ELECTRICAL   PRACTICE 

Efficiency  of  Electrical  Machinery  and  Processes — Transmission  of 
Electrical  Power — Efficiency:  Effect  of  Voltage — Relation  of 
Size  of  Conductor  to  Voltage  of  Transmission — Feeders — 
Three-wire  System — Kirchhoff's  Laws — Current  Distribution 
in  Parallel  Combinations  of  Battery  Cells  or  Generators — 
Stray  Power  Losses — Commercial  Efficiency  of  Generators 
and  Motors — Electrical  Efficiency  of  Generators — Mechanical 
Efficiency  of  Motors 231 

CHAPTER  X 
INDUCTANCE 

Mutual  Inductance — Cause  of;  Effect  of;  Lenz's  Law — Induction 
Coils;  Jump  Spark;  Ruhmkorff — Transformers — Self  Induct- 
ance; Cause  and  Effect  of — Induction  Coil;  Make  and  Break 


X  TABLE  OF  CONTENTS 

PA6E 

— Inductance,  a  Property  of  the  Circuit;  Unit  of  Inductance, 
the  Henry — Computation  of  Self  Inductance;  of  Mutual 
Inductance;  of  Inductance  in  Transmission  Lines — Effect  of 
Inductance  in  Alternating  Current  Circuits 268 

CHAPTER  XI 
CAPACITY 

Capacity:  The  Elasticity  of  an  Electric  Circuit — Farad  and  Micro- 
farad, the  Unit  of  Capacity — Relation  of  Charge,  Voltage  and 
Capacity — Positive  and  Negative  Electricity — Bound  and 
Free  Charges — Condensers:  Equation  for  Capacity  of — Di- 
electric Power — Dielectric  Strength — Capacity  of  Cables; 
Equation  for — Measurement  of  Capacity;  by  Direct  Deflec- 
tion of  Ballistic  Galvanometer;  Bridge  Method — Locating  a 
Break  in  a  Cable — Capacity  of  Condensers  Joined  in  Parallel 
and  in  Series 300 

CHAPTER  XII 
ELECTROCHEMISTRY 

Primary  and  Secondary  Cells — E.M.F.  Dependent  on  Materials — 
Chemical  Action  of  Simple  Cell — Electrolysis — Electrochem- 
ical Equivalents — Electroplating  and  Electrotyping — Storage 
Cells;  Theory,  Construction  of  Plante,  Faure,  and  Edison 
Types— Use  and  Maintenance — Advantages  and  Disadvan- 
tages   324 

CHAPTER   XIII 
PHOTOMETRY  AND  ELECTRIC  ILLUMINATION 

Illumination;  Distribution,  Color,  Intensity — Law  of  Inverse 
Squares — Photometry,  Candle  Power;  Horizontal  and  Spher- 
ical— Foot-Candle — Use  of  Sharp-Millar  Illuminometer — Arc 
Lamp;  Ballast  and  Regulating  Resistances — Flaming  or 
Luminous  Arcs — Incandescent  Lamps,  Carbon,  Tantalum, 
Tungsten,  Nernst — Life;  Effect  of  Overburning;  of  Under- 
burning — Mercury  Vapor  Lamp — Moore  Tube 363 


TABLE  OF  CONTENTS  xi 

CHAPTER   XTV 
ELECTRICAL  MEASURING  INSTRUMENTS 

PAGE 

Galvanometer;  D' Arson  val  and  Thomson  Types — Deflecting 
Force;  Control;  Damping — Sensibility — Shunts;  Ayrton 
Universal  Shunt — Series  Resistance;  Megohm — Ballistic 
Galvanometer — Thermal  Effects — Ammeters;  Resistance  of 
— Types:  Solenoidal,  Hot-Wire;  Permanent  Magnets,  Two 
Coil  or  Electro-Dynamometer — Voltmeters,  Resistance  of — 
Types;  Solenoidal;  Hot-Wire,  Permanent  Magnet,  Two  Coil, 
Electro-Static — Wattmeter — Construction  of  Weston  Type — 
Thomson  Integrating  Wattmeter  or  Watt-hour  Meter — 
Potentiometer;  Theory,  Construction  and  Use — Voltameter.  387 

CHAPTER  XV 
ALTERNATING  CURRENTS 

Definition:  Cycle;  Frequency;  Phase;  A.C.  Generator;  Vector 
Diagrams,  Average,  Maximum,  Effective  and  Instantaneous 
Values  of  E.M.F.  and  Current — Computation  of  Current; 
Current  and  Voltage  in  Phase ;  Leading  Current ;  Lagging  Cur- 
rent— Causes  and  Effects  of  Lead  and  Lag — Reactance, 
Inductance,  and  Capacity — Computation  of  Reactance; 
Impedance;  Computation  of — Ohm's  Law  for  A.C.  Circuits—- 
Series and  Parallel  Circuits — Power  in  A.C.  Circuits — General 
Law  for  A.C.  Circuits;  when  Current  and  Voltage  are  in 
Phase;  when  out  of  Phase — Power  Factor — Effect  oi  Induct- 
ance and  Capacity  on  Power  Factor — Use  of  Ammeter  and 
Voltmeter  for  A.C.  Power  Measurements — Use  of  Wattmeter.  430 

APPENDIX 

Useful  Numbers — Metric-English  Equivalents — Mechanical  Equiv- 
alents of  Heat — Significant  Figures — Plotting  of  Curves — 
Equation  of  Straight  Line — Simple  Trigonometric  Functions 
— Typical  Values  of  B  and  H  for  Different  Irons — Resis- 
tivity and  Temperature  Coefficients  of  Various  Metals  and 
Commercial  Alloys — Wire  Tables  for  Copper — Safe  Carrying 
Capacity  of  Copper  Wires — Equivalent  Values  of  Electrical, 
Mechanical  and  Heat  Units...  .  525 


ELEMENTS  OF  ELECTRICITY 


CHAPTER  I 
MAGNETS    AND    MAGNETISM 

Definition  of  Magnet — Poles — Magnetic  Lines — Flux — Field  Intensity — 
Nature  of  Magnetic  Lines — Attraction  and  Repulsion  of  Magnets 
— Permeability — Induction — Magnetic  Screens — Compass — Unit 
Poles — Measure  of  Mutual  Action  between  Two  Magnets — 
Mathematical  Relation  between  Unit  Pole  and  Force  Line — 
Two  Equations  for  Magnetic  Force — Ring  Magnets — Conse- 
quent Poles — Permanent  Magnets — Magnetic  Molecule  Theory. 

IN  machines  for  the  generation  and  use  of  electricity  at 
the  present  time,  some  form  or  other  of  magnet  is  required. 
A  thorough  knowledge  of  magnets  and  the  principles  of 
magnetism  is  therefore  necessary,  in  order  to  understand 
the  construction  and  operation  of  generators,  motors,  and 
other  electrical  appliances.  Accordingly,  the  subject  will 
be  presented  only  as  it  bears  on  electric  machinery,  and 
no  attempt  will  be  made  to  study  it  in  its  other  aspects. 

1.  Magnets  and  Magnetism.  It  was  known  to  the 
ancient  Greeks  that  a  "lodestone"  or  natural  magnet 
possessed  the  power  of  attracting  bits  of  iron  to  itself, 
and  even  could  impart  this  attractive  power  to  the  iron 
pieces  themselves,  when  they  were  rubbed  on  it. 

This  property  of  attracting  iron  and  steel  is  called  MAG- 
NETISM, and  a  body  possessing  it  is  called  a  MAGNET. 

Under  the  action  of  a  sufficient  magnetizing  force,  probably 
all  materials  possess  this  property  to  some  degree  at  least, 
though  iron  and  steel  possess  it  in  so  much  greater  degree 


2        ,    t  ,     .    ,     ELEMENTS  OF   ELECTRICITY 

than  all  other  materials  that  they  are  generally  referred 
to  as  the  MAGNETIC  substances,  in  distinction  from  all  other 
materials.  And  even  iron  and  steel  differ  in  their  magnetic 
qualities.  Steel,  on  being  magnetized,  "with  proper  treat- 
ment remains  magnetized  indefinitely,  whereas  iron,  and 
especially  soft  iron,  loses  its  magnetism  as  soon  as  the 
magnetizing  force  is  removed.  Thus  PERMANENT  MAGNETS 
are  made  of  hard  steel,  and  TEMPORARY  MAGNETS  of  soft  iron 
or  annealed  steel. 

2.  Poles.     Certain  parts,  only,  of  a  magnet  possess  this 
power  of  attracting  iron.     These  parts  are  called  the  POLES, 
either  NORTH  OT  SOUTH,  according  to  a  distinction  explained 
later.     A  line  drawn  across  the  magnet  half  way  between 
the  poles  is  called  the  EQUATOR. 

3.  Magnetic   Lines.     The  nature  of  magnetism  is  that  of 
a  stream  or  CURRENT.     This  current  of  magnetism  is  repre- 
sented by  lines,  called  MAGNETIC  LINES  OF  FORCE,  which 


FIG.  1. — Bar  magnet. 

always  flow  out  of  the  North  pole,  around  into  the  South 
pole,  and  back  through  the  magnet  to  the  North  pole  again, 
thus  forming  a  complete  circuit.  A  NORTH  POLE,  then, 
may  be  defined  as  that  portion  of  a  magnet  out  of  which  the 
lines  of  force  flow;  a  SOUTH  POLE,  that  portion  into  which 
they  flow.  For  convenience  each  pole  is  often  considered 
to  be  concentrated  at  a  point  near  the  end.  A  bar  magnet 
is  represented  as  in  Fig.  1.  N  =  North  pole;  S  =  South  pole; 
E  =  equator.  Each  magnetic  line  forms  a  complete  loop  or 
CIRCUIT,  called  a  MAGNETIC  CIRCUIT.  The  total  number 
of  lines  threading  the  circuit  makes  up  the  magnetic  FLUX 


MAGNETS  AND   MAGNETISM  3 

and  is  usually  represented  in  formulas  by  the  symbol  (f>. 
Thus  when  the  expression,  <£=  10,000  is  written,  it  means 
that  the  total  number  of  lines  of  force  passing  through 
a  given  area  is  10,000.  These  lines  of  force  are  as  true 
force  lines  as  the  lines  drawn  to  represent  the  tension  in 
ropes,  rods,  etc.,  and  can  be  combined  into  resultants  and 
resolved  into  components  as  well  as  any  other  force  lines. 

The  fact  that  the  lines  actually  run  through  this  magnet 
and  form  a  complete  loop,  and  do  not  merely  start  at  the 
N  pole  and  end  at  the  S  pole,  is  proved  by  breaking  a  magnet 
into  several  pieces.  Each  piece  becomes  a  separate  magnet 
with  a  N  pole  and  S  pole  of  its  own.  See  Fig.  2. 


Fia.  2.     A  magnet  broken  into  four  pieces,  each  of  which  has  its  own  N.  and  S.  pole. 

For  convenience,  the  poles  are  sometimes  considered 
to  be  points  within  the  magnet,  toward  which  all  the  lines 
seem  to  converge. 

4.  Magnetic  Field.  The  space  outside  the  magnet 
occupied  by  the  magnetic  lines  is  called  the  FIELD.  The 
number  of  lines  per  sq.cm.  (of  surface  at  right  angles  to  lines) 
is  called  the  FIELD  INTENSITY,  or  Field  Strength. 

A  field  intensity  of  1  line  per  sq.cm.  is  called  a  GAUSS.  Thus 
10  gausses  simply  means  10  lines  per  sq.cm.  The  symbol  for 
field  intensity  is  H.  To  find  the  total  flux  (j>  in  a  given  area, 
we  have  merely  to  multiply  the  field  intensity  H  by  the  area  A . 


4  ELEMENTS  OP  ELECTRICITY 

We  may  express  this  relation  by  the  equation 

(f>=AH 
which  means 

<£  (lines)  =  ;!  (sq.cms.)  X//  (gausses  or  lines  per  sq.cm.). 

Example.  What  is  the  total  flux  flowing  in  a  magnetic  field 
of  80  sq.cms.,  if  the  average  field  strength  is  2  kilogausses  (i.e. 
2000  gausses)? 


<j>  =  80(sq.cms.)  X  2000  (gausses)  =  160,000  lines. 

Problem  1-1.  The  area  of  a  pole  face  of  a  motor  is  600  sq.cms. 
The  average  field  intensity  of  magnetic  field  for  this  surface  is 
5000  gausses.  What  is  the  total  number  of  magnetic  lines  com- 
ing out  of  the  pole  face? 

Problem  2-1.  What  area  must  the  pole  face  of  a  generator 
have,  in  order  that  the  total  magnetic  flux  may  be  10,000,000 
lines?  #=8000  gausses. 

Problem  3-1.  What  is  the  intensity  at  a  point  where  400 
sq.cms.  have  a  flux  of  2,000,000  magnetic  lines? 

5.  Distribution  of  Field  Intensity.  That  part  of  a 
magnet  will  attract  iron  with  the  greatest  force  where  the 
field  intensity  is  greatest,  that  is,  where  the  lines  are  flowing 
either  out  of  or  into  the  magnet  in  the  greatest  number  per 
sq.cm.  (The  attraction  is  proportional  to  the  square  of 
the  number  of  lines  per  sq.cm.) 

The  analogy  to  a  stream  of  water  is  apparent;  wherever 
the  stream  spreads  out,  the  current  is  weak,  and  wherever 
it  is  contracted,  the  current  is  strong.  Thus  wherever  the 
magnetic  current  spreads  out  and  covers  a  large  area,  it 
is  weak,  and  there  are  few  lines  per  sq.  cm.,  but  wherever 
it  is  contracted  into  a  small  space,  as  near  the  magnet, 
there  the  current  is  strong  and  the  number  of  lines  per 
sq.cm.  is  great.  Thus  the  intensity  of  the  field  is  greatest 
near  the  poles  and  grows  less  as  we  go  away  from  them. 
Fig.  3  illustrates  this. 

A  represents  a  square  centimeter.  Notice  that  if  this  were 
moved  nearer  the  magnet,  a  larger  number  of  lines  would  pass 


MAGNETS  AND  MAGNETISM 


through  it.     If  it  were  placed  at  one-half  this  distance,  the  number 

of  lines  passing  through  it  would  be  not  merely  twice  as  great,  but 

four  times  as  great.     At   one-third 

the  distance,  the  number  of  lines  per 

sq.cm.  becomes  nine  times  as  great, 

etc. 

This  may  be  stated  as  a  general 
law  as  follows : 

The  intensity  at  any  point  in  the 
field  of  an  isolated  magnetic  pole 
varies  inversely  as  the  square  of  the 
distance  of  that  point  from  the  pole. 

Or  in  an  equation ;    //oe  -^-. 

6.  Nature  of   Magnetic  Lines 
of     Force.       (1)   Magnetic  lines 
represent  a  tension    along   their 
length   and    tend   to  shorten  as 
stretched  rubber  bands  do. 

(2)  They  also  exert  a  lateral  crowding  effect  on  one  another, 
tending  to  push  one  another  sideways. 

By  means  of  these  two  Jacts  we  can  explain  many  magnetic 
phenomena.  "" \*.  (^NAJ-* 

7.  Attraction    and    Repulsion.     When  we  place  the  N 
pole  of  one  magnet  near  the  S  pole  of  another,  the  magnets 
attract  one  another.     Fig.  4  shows  this  action.     The  lines 


FIG.  3. — Intensity  of  magnetic 
field! 


FIG.  4. — Unlike  poles  attract  each  other. 

coming  out  of  B  enter  the  S  pole  of  A,  go  through  this, 
and  emerging  from  the  N  pole  of  A,  return  to  the  S  pole 
of  B.  The  tension  in  the  lines  thus  threading  the  two 


6 


ELEMENTS   OF  ELECTRICITY 


magnets  tends  to  pull  the  magnets  together,  as  so  many 
loops  of  rubber  bands  would. 

If  we  place  the  north  ends  near  each  other  as  in  Fig.  5, 
the  lines  coming  out  of  the  'N  poles  turn  one  another  aside 


FIG.  5. — Like  poles  repel  each  other. 

and  their  lateral  crowding  pressure  tends  to  push  the 
magnets  apart. 

The  rule,  then,  is  that  unlike  poles  attract,  while  like 
poles  repel  one  another. 

If  we  place  the  magnets  under  a  piece  of  glass,  on  which 
iron  filings  have  been  sprinkled,  the  filings  will  arrange  them- 
selves along  the  lines  of  force  as  shown  in  Figs.  6  and  7.  This 
is,  perhaps,  the  truest  picture  of  an  external  magnetic  field. 

8.  Permeability.  Magnetic  lines,  as  previously  indicated, 
may  be  set  up  in  any  substance,  though  much  more  easily 
in  some  substances  than  in  others. 


FIG.  6. — Shape  of  field  between  like 
poles  shown  by  iron  filings 


FIG.  7. — Shape  of  field  between  unlike 
poles  shown  by  iron  filings. 


Since  we  can  set  up  1  line  of  force  in  1  cu.cm.  of  air  by  1 
unit  of  magnetizing  force,  we  say  that  air  has  a  PERMEABILT 
ITY  of  1.  The  same  force  might  set  up  about  1500  lines 
in  the  same  amount  of  wrought  iron  and  1000  lines  in 


MAGNETS  AND   MAGNETISM  7 

steel.  Therefore  we  say  the  permeability  of  wrought  iron 
is  1500  and  of  steel  is  1000,  though  it  varies  greatly  in 
different  specimens;  and  even  the  same  specimen  will  vary 
as  to  its  permeability,  as  explained  in  Chapter  VI. 

We  may  then  define  the  permeability  of  a  substance  as 
the  ratio  of  the  number  of  magnetic  lines  set  up  in  a  unit 
volume  of  that  substance,  to  the  number  set  up  in  the  same 
amount  of  air  by  the  same  magnetizing  force. 

9.  Induction.  The  attraction  of  pieces  of  iron  or  steel 
to  a  magnet  is  a  direct  result  of  the  high  permeability  of 
iron  and  steel.  When  a  piece  of  soft  iron,  for  instance,  is 
placed  in  a  magnetic  field,  its  permeability  being  so  much 
greater  than  that  of  the  surrounding  air,  a  larger  number 
of  magnetic  lines  is  immediately  set  up  in  it  than  in  the 
air.  The  action  is  as  though  the  lines  left  the  air  and 
crowded  into  the  iron.  The  piece  of  iron  is  then  said  to 
become  a  magnet  by  INDUCTION.  This  explains  the 
action  of  iron  filings  when  scattered  in  a  magnetic  field. 
Each  particle  of  iron  becomes  a  miniature  magnet  and  acts 
like  a  compass  needle. 


Fia.  8. — Piece  of  soft  iron,  A ,  has  become  magnetized  by  induction  and  is  attracted 
to  the  magnet,  M. 

MAGNETIC  INDUCTION  is,  then,  the  setting  up  in  a  magnetic 
substance  of  a  greater  number  of  lines  than  would  be  set  up 
in  air  by  the  same  cause. 

Fig.  8  shows  this  process.      A  is  a  small  piece  of  iron 


8  ELEMENTS   OF  ELECTRICITY 

which  was  not  magnetized  before  it  was  placed  in  the 
magnetic  field  of  the  magnet  M.  The  field  is  distorted  by 
the  lines  apparently  trying  to  leave  the  air  and  go  through 
the  iron,  which  now  has  a  N  and  S  pole  of  its  own,  since 
lines  of  force  enter  one  end  and  leave  the  other.  We  have, 
then,  two  magnets  with  their  unlike  poles  near  each  other, 
and  they  therefore  attract  one  another.  This  explains  why 
any  piece  of  soft  iron  placed  near  a  magnet  is  always 
attracted  to  it.  The  strength  of  the  INDUCED  MAGNET 
depends  upon  the  permeability  of  the  iron  and  the  strength 
H  of  the  magnetic  field  in  which  it  happens  to  lie.  The 
greater  the  field  strength  H,  the  stronger  magnet  it  makes 
of  a  piece  of  iron.  For  this  reason  the  field  strength  H 
is  sometimes  called  the  MAGNETIZING  FORCE  of  the 
field.  Since  the  field  strength  H  is  greater  nearer  the 
poles,  so  the  magnetizing  force  on  a  piece  of  iron  is  also 
greater,  the  nearer  the  poles  the  piece  of  iron  is  placed. 

This  explains  why  a 
piece  of  iron  near  a  pole 
is  attracted  by  so  much 
greater  force  than  wh'en  it 
is  further  away  from  the 
pole. 

10.  Magnetic  Screens. 
"•  > •     This  inductive  property  of 

FIG.  9. — Piece  of  iron,  B,  acts  as  a  screen      iron    is   Used   also,  whenever 
and  conducts  lines  around  space  A.  . 

it  is  desired  to  cut  the  lines 

of  force  out  of  any  part  of  a  magnetic  field.  A  shield  of  soft 
iron  is  put  in  the  field  as  in  Fig.  9.  Tlie  iron  screen  B,  since 
it  has  a  much  higher  permeability  than  air,  takes  up  nearly 
all  the  lines  in  the  field  H  and  conducts  them  around  the 
space  A.  Instruments  are  sometimes  thus  shielded  from 
t'he  effects  of  a  magnetic  field.  There  is  no  Magnetic 
Insulator,  that  is,  a  material  which  will  stop  the  lines,  or 
one  whose  permeability  is  zero.  So  the  above  method 
of  conducting  as  many  of  the  lines  as  is  possible  around  a 


MAGNETS  AND    MAGNETISM  9 

certain  place,  is  resorted  to,  when  it  is  desired  to  free  it  from 
magnetic  lines. 

11.  Flux  Density.  The  number  of  magnetic  lines  per  sq.cm. 
in  any  substance  is  called  the  FLUX  DENSITY  in  that  substance. 

The  flux  density  in  air  is  always  //,  which  we  have  seen  is  also  the 
magnetizing  force.  Note  carefully  these  two  uses  for  the  letter  H: 

(1)  Flux  density  in  air. 

(2)  Magnetizing  force. 

The  flux  density  in  any  material  other  than  air  is  represented 
by  the  letter  B,  and  under  the  influence  of  the  same  magnetiz- 
ing force  is  as  much  greater  than  H,  as  the  permeability  of  the 
material  is  greater  than  the  permeability  of  air. 

Thus,  in  Fig.  8,  the  piece  of  soft  iron  A  was  placed  in  the 
magnetic  field  of  the  magnet  M.  The  flux  density  of  the  air 
at  that  place,  before  the  iron  A  was  placed  in  it,  was  H,  but 
when  the  iron  was  placed  in  the  field  ,the  flux  density  within  the 
iron  immediately  became  B  which  is  much  greater  than  H, 
because  the  permeability  of  the  iron  is  greater  than  air. 

The  PERMEABILITY  of  a  substance  may  thus  be  defined  as  the 
ratio  of  B  to  H  ,  that  is,  the  ratio  of  the  flux  density  to  the  magnetiz- 
ing force, 

The  Greek  letter  ft  is  used  to  express  the  value  of  the  permeability. 
Thus  we  may  write  the  equation 

B 

f"J! 

where    ft  =  Permeability  (pure  number)  ; 
J5  =  Flux  density  in  gausses', 
H  =  Magnetizing  force,  gausses. 

Since  in  the  air  the  magnetizing  force  is  H  and  the  flux  density 
is  also  H,  the  equation  for  the  permeability  of  air  becomes, 

,-§-, 

Example.  The  piece^  of  iron,  A,  Fig.  8,  has  a  permeability 
of  900.  It  is  placed  at  a  point  in  the  magnet's  field  where  the 
intensity  is  4  gausses.  What  is  the  flux  density  in  the  iron? 

B 


=  900X4=3600  gausses. 

Problem  4-1.  How  many  lines  of  force  per  sq.cm.  will  be  set 
up  in  a  piece  of  steel  by  a  magnetizing  force  of  18  gausses?  The 
permeability  of  the  steel  is  1200. 


10  ELEMENTS  OF  ELECTRICITY 

Problem  5-1.  It  is  desired  to  set  up  400,000  magnetic  HRCS 
of  force  in  a  piece  of  iron,  the  cross-section  of  which  is  8  cms.  X 10 
cms.  If  the  permeability  of  the  iron  is  750,  what  magnetizing 
force  is  required? 

12.  The  Compass.  If  a  magnet,  which  is  free  to  turn 
on  its  axis,  is  placed  in  a  magnetic  field,  it  will  take  a  posi- 
tion such  that  the  lines  of  force  within  it  run  in  the  same 
direction  as  those  in  the  magnetic  field  in  which  it  is  placed. 

This  is  the  rule  for  the  action  of  the  compass.  A  com- 
pass is  merely  a  bar  magnet  suspended  so  that  it  turns 
freely.  The  earth  is  a  large  magnetic  field  with  the  lines 
running  from  south  to  north.  Thus  a  compass  takes  a 
position  parallel  to  these  lines,  or  nearly  south  and  north. 

Of  course,  according  to  the  definition  we  have  given  for 
north  and  south  poles,  the  North  Magnetic  Pole  is  near  the 
South  Geographical  Pole  and  vice  versa.  It  causes  no  con- 
fusion to  say  that  the  compass  points  nearly  to  the  North 
Pole. 

The  place  to  which  the  north  end  of  a  compass  points  is  not 
the  geographical  North  Pole  of  the  earth,  but  a  spot  in  Boothia, 
a  peninsula  in  the  northern  part  of  Canada.  At  this  spot,  the 
earth's  magnetic  lines  enter  the  earth,  run  through  the  earth, 
emerge  near  the  geographical  South  Pole,  and  flow  north  along 
the  earth's  surface.  This  gives  us  on  the  earth's  surface  a  prac- 
tically uniform  magnetic  field  flowing  from  south  to  north,  called 
the  Earth's  Magnetic  Field. 

Practically,  we  are  not  so  much  interested  in  the  North  and 
South  Pole  as  in  the  exact  direction  of  the  lines  at  different  parts 
of  the  earth's  surface;  especially  that  part  which  is  out  of  sight 
of  land,  and  where  the  compass  must  be  depended  upon  for  direc- 
tion. In  order  to  map  very  completely  the  earth's  magnetic 
field,  The  Carnegie  Institute  has  had  built  at  Brooklyn  a  special 
ship  called  the  "Carnegie."  This  vessel  is  remarkable  in  that  it 
is  constructed  almost  entirely  of  non-magnetic  materials.  With 
the  delicate  instruments  which  can  be  used  on  the  "Carnegie," 
an  accurate  magnetic  survey  of  the  important  portions  of  the 
earth's  surface  is  being  made. 

For  a  discussion  of  other  magnetic  phenomena,  see  any  encyclo- 
pedia. Some  special  topics  of  interest  are :  History  of  the  compass; 
determination  of  the  inclination  and  declination  of  the  magnetic 
needle;  intensity  of  the  earth's  field;  paramagnetic  and  diamag- 


MAGNETS    AND   MAGNETISM  11 

netic  materials;   magnetization  of  gases  and  fluids;   effects  of  the 
medium  in  which  the  magnetic  field  lies. 

All  magnets  which  we  use  are  in  the  earth's  magnetic 
field.  Thus  the  field  about  a  magnet  is  really  the  resultant 
field  of  the  earth's  field  and  the  magnet's  field. 

13.  Unit  Pole  and  Field  Strength.  It  has  already 
been  stated  that  the  intensity,  or  strength,  at  any  point  in 
a  magnetic  field  is  measured  by  the  number  of  lines  passing 
through  a  square  centimeter  of  it. 

Another  way  to  measure  the  strength  at  a  certain  point 
in  a  magnetic  field,  is  to  measure  the  amount  of  force  that 
it  exerts  on  a  unit  magnetic  pole  placed  at  that  point. 
When  the  field  exerts  a  force  of  one  dyne  on  a  unit  pole, 
the  intensity  or  strength  of  that  part  of  the  field  is  s'aid  to 
be  1  line  per  sq.cm.,  or  better,  1  gauss.  The  two  methods 
of  stating  the  field  strength  give  the  same  numerical  value, 
1  gauss,  2  gausses,  etc.,  which  is  represented  by  the  sym- 
bol H. 

A  UNIT  POLE  is  a  pole  of  such  strength  that,  if  placed 
a  centimeter  away  in  the  air  from  a  like  pole,  it  will  repel 
it  with  a  force  of  one  dyne.  An  isolated  N  pole  is  not  a 
physical  possibility,  since  every  magnet  must  have  a  south 
pole  for  every  north  pole.  We  may,  however,  consider 
how  such  a  pole  would  be  acted  upon  when  placed  in  a 
magnetic  field. 

If  a  unit  north  pole  were  placed  near  the  north  end  of 
a  bar  magnet  as  in  Fig.  1,  it  would  be  repelled  by  the 
north  end  of  the  magnet  and  would  travel  to  the  south  end, 
not  by  the  shortest  path,  but  along  any  line  of  force 
on  which  it  might  happen  to  lie.  Therefore  a  magnetic 
line  of  force  represents  the  line  along  which  a  north  pole 
is  repelled  or  a  south  pole  attracted. 

When  the  field  is  strong  enough  tox  repel  a  unit  north 
pole  along  a  line  with  the  force  of  one  dyne,  we  say  the  field 
has  a  strength  of  one  dyne,  or  of  1  gauss,  i.e.,  1  line  per 
sq.cm. 


12  ELEMENTS  OF  ELECTRICITY 

We  may,  then,  also  say  that  a  unit  pole  is  a  pole  of  such 
strength  that  when  placed  in  a  field  of  one  gauss,  it  is  acted 
upon  by  a  force  of  one  dyne. 

14.  Effect  of  a  Magnetic  Field  upon  Another  Magnet.  Since 
a  magnet  of  unit  pole  strength  is  acted  upon  by  a  force  of  one 
dyne  when  placed  in  a  unit  field,  it  follows  that  if  the  pole  strength 
is  doubled,  the  action  of  the  unit  field  on  the  magnet  would  be 
doubled.  Or  if  we  increased  the  field  strength,  its  action  on  the 
magnet  would  be  proportionally  increased. 

Thus  we  may  compute  the  action  of  a  magnet  placed  in  the  field 
of  another  magnet,  by  multiplying  the  pole  strength  of  the  magnet, 
by  the  strength  of  that  part  of  the  field  in  which  it  is  placed.  As 
an  equation,  this  may  be  stated  as  follows: 

F  =  mH. 

where  F=  action  of  field  upon  magnet  in  dynes; 
m  =  strength  of  magnet  in  unit  piles', 
H  =  strength  of  magnetic  field  in  gdusses. 

Example.  A  magnet  of  400  unit  poles  is  placed  in  a  magnetic 
field  of  5  kilogausses.  What  force  is  exerted  upon  the  magnet? 


F  =  400(unit  poles)  X  5000  (gausses) 
=  2.000,000  dynes 
=  2040  grams  =  4.  5  Ibs. 

Problem  6-1.  A  magnet,  when  placed  in  a  magnetic  field  of 
200  gausses  is  attracted  by  a  force  of  .3  Ib.  How  many  unit 
poles  has  the  magnet? 

Problem  7-1.  If  a  magnet  of  800  units  pole  strength  were 
placed  in  a  field  of  1.4  kilogausses,  what  force  would  be  exerted 
on  the  magnet? 

15.  Action  of  the  Compass.  Since  every  magnet  has 
a  north  end  and  south  end,  there  is  always  a  double  action 
upon  it  when  placed  in  a  magnetic  field.  The  north  end 
will  tend  to  move  along  the  lines  of  force  in  the  direction 
of  the  field,  and  the  south  end,  in  the  opposite  direction. 
If  the  magnet  is  free  to  turn,  as  a  compass  is,  it  will  turn 
until  it  lies  along  a  line  of  force. 

This  explains  why  a  compass  always  takes  a  position 
guch  that  the  lines  of  force  within  it  are  in  the  same  direc- 


MAGNETS   AND   MAGNETISM 


13 


tion  as  the  field  in  which  it  lies.     It  always  tends  to  lie  along 

a  line  of  force,  and  if  placed 

otherwise,   there  will  be  a 

"  force   couple  "    acting   on 

it  to  turn  it  into  that  posi- 

tion as  in   Fig.    10.     Let  a 

compass,  pivoted  at  A,  with 

pole  strength  m  be  placed 

in  a  position  NS  in  a  field 

whose  direction  is  indicated 

by  arrows  HH.     The  north 

pole  N  would  be    repelled 

along   the  direction    of    H 

by  a  force  mH,  and  the  south  pole  S  would  be  attracted 

in  the  opposite  direction  by  a  force  m'H  equal  to  mH,  and 

the    compass    would    be   turned  around  into  the  position 

Nr  S',  parallel  to  force  lines  of  field  in  which  it  lies. 

16.  Mutual  Action  between  Two  Magnets.  It  has  been  found 
by  experiment,  that  the  force  with  which  the  poles  of  two  magnets 
attract  or  repel  each  other,  in  the  air,  is  equal  to  the  product 
of  their  pole  strengths,  divided  by  the  square  of  the  distance 
between  them,  or; 


FIG.  10. — Action  of  magnetic  field,  H, 
on  compass,  A. 


where  F  =  force  in  dynes  ; 

m  =  strength  in  unit  poles  (of  one  pole)  ; 
m'  =  strength  in  unit  poles  (of  other  pole)  ; 
d  =  distance  between  poles  in  cms. 

Example.  If  the  N  pole  of  a  magnet  of  10  units  strength  be 
placed  4  cms.  from  the  N  pole  of  another  magnet  of  24  units  pole 
strength,  with  what  force  will  they  repel  each  other? 

(Consider  the  magnets  of  sufficient  length  that  the  S  poles  do 
not  affect  the  action  of  the  N  poles). 


m=10; 


10X24 


=15  yneSf 


14  ELEMENTS  OF  ELECTRICITY 

Problem  8-1.  How  far  from  a  pole  of  400  units  pole  strength 
must  another  pole  of  3000  units  pole  strength  be  placed  in  order 
that  the  force  between  them  may  be  .2  lb.? 

Problem  9-1.  Two  poles,  2.4  cms.  apart,  attract  each  other 
with  a  force  of  4940  dynes.  One  has  a  strength  of  2430  unit  poles. 
What  strength  has  the  other?  . 

17.  Relation  between  Unit  Pole  and  Force  Lines.  There 
are,  then,  two  ways  of  indicating  the  pole  strength  of  a  magnet; 
either  : 

(1)  By  the  number  of  unit  poles  it  contains,  or, 

(2)  By  the  number  of  lines  of  force  coming  out  of  it. 

It  is  necessary  often  to  reduce  one  set  of  units  to  the  other. 
This  can  be  done  as  follows: 

A  unit  pole  may  be  thought  of  as  a  point  which  sends  out  one 
line  of  force  to  every  sq.cm.  of  surface  situated  1  cm.  distant  from 
the  pole.  The  shape  of  a  surface,  everywhere  1  cm.  distant  from 
a  point,  is  a  sphere  of  1  cm.  radius. 

Area  of  a  sphere  =  4^r2 

where  r  =  radius  of  sphere  ; 

In  this  case  we  have  said  r  =  1  cm. 


7r  sq.cms. 

Since  there  is  one  line  to  each  sq.cm.  of  surface,  there  must  be 
4?r  lines  to  the  complete  surface. 

Therefore  each  unit  pole  is  conceived  to  have  a  total  of  4?r  lines 
of  force  coming  out  of  or  going  into  it. 

A  magnet  with  the  strength  of  24  unit  poles  would  therefore 
have  24X4*  or  302  lines  flowing  out  of  the  N  pole  around  into 
the  S  pole.  At  a  distance  of  1  cm.  from  the  pole,  however,  there 
would  be  a  field  intensity  of  24  gausses,  or  24  lines  per  sq.cm. 

18.  Relation  between  Equations  for  Magnetic  Force.  We 
have  then  two  equations  for  magnetic  force  F: 

Force  between  two  magnets, 


Force  on  magnet  in  magnetic  field  ; 

F=mH  ..........     (2) 

It  is  well  to  consider  the  relation  between  these  two  equations, 
(1st)  Mathematically,  and  (2d)  Analytically. 

(1st)  Mathematically.  We  have  seen  that  the  field  strength  H 
at  1  cm.  from  a  unit  pole  is  1  gauss,  or  1  line  per  sq.cm.  The 
field  strength  at  1  cm.  distance  from  pole  (m)  of  64  units  pole 
strength  would  be  64  gausses,  'or  64  lines  per  sq.cm. 


MAGNETS  AND  MAGNETISM  15 

Also  we  have  seen  that  the  field   strength   H  varies  inversely 
as  the  square  of  the  distance  from  the  pole  m'  '. 
Thus  we  may  write  the  equation 


where  H  =  field  strength  in  gausses; 
mf  =  pole  strength  in  unit  poles  ; 
d  =  distance  in  centimeters. 

If  now  we  substitute  this  value  for  H  in  Eq.  (2),  F=mXH,  we 
get, 


But  this  is  the  same  as  Eq.  (1)  for  the  force  exerted  by  one  magnet 
on  another. 

Thus  we  see  that  the  force  exerted  by  one  magnet  on  another 
is  merely  the  force  exerted  by  its  magnetic  field  on  that  magnet, 
and  can  readily  be  found  by  the  general  equation  for  the  action 
of  a  magnetic  field  on  a  magnet  placed  in  it,  F=mH. 

(2d)  Analytically.  Consider  Fig.  11.  Let  m  =  pole  of  64  units 
pole  strength.  There  is,  then,  a  field  intensity  of  64  gausses  at 
a  distance  of  1  cm.  from  the  pole;  that  is,  64  lines  emerge  from 
the  pole  for  every  sq.cm.  of  the  surface  on  a  globe  1  cm.  from  the 
pole.  Thus  the  field  intensity  along  the  surface  A  A  1  cm.  from 
the  pole  =  64  gausses.  Along  a  surface  BB,  2  cms.  from  the 
pole,  the  field  intensity  would  not  be  one-half  as  great,  but  only 
one-quarter  as  great;  that  is,  ^r,  or  16  gausses.  At  3  cms.  distant, 
the  field  would  become  only  i  as  great,  or  7.1  gausses.  Thus  while 
there  are  64  lines  per  sq.cm.  at  1 

cm.  distance,  there  are  only  16  lines  ~~"^N 

per  sq.cm.  at  2  cms.  and  7.1  lines  per  \ 

sq.cm.  at  3  cms.  distance.     This  fol-  A~^X       \ 

lows  from  the  fact  that  the  strength 
of  the  field  about  a  single  magnetic 
pole  varies  inversely  as  the  square  of 
the  distance  from  the  pole. 

Suppose  now  a  magnetic  pole  mf 
of  10  units  pole  strength  be  placed 
1  cm.  from  m,  of  64  units.  It  would 
be  in  a  field  of  64  gausses  as  explained 

above.     The  force  exerted  upon  it  would  then  be  //Xw  =  64X10  = 
640  dynes. 

Also  according  to  the  equation  (1)  the  force  would  be 

mm'     64X10 


16  ELEMENTS   OF  ELECTRICITY 

Suppose  this  magnet  of  10  units  pole  strength  be  placed  2  cms. 
away  from  m.  It  would  now  be  in  a  field  of  16  gausses,  as  seen 
above,  and  the  action  would  be  HXm  =  16X10  =  160  dynes. 

Or  by  other  method  : 

mm'     64X10 

--==-~=  160  dynes. 


Thus  it  is  seen  that  the  two  equations  F  =  —^~   and  F  =  Hm 

are  identical  and  give  the  same  result. 

Therefore,  if  we  know  the  number  of  unit  poles  a  magnet  con- 
tains and  the  strength  of  the  field  in  which  it  lies,  and  we  wish  to 
know  the  force  exerted  on  the  magnet  we  have  merely  to  use  the 
equation 

F  =  mH. 

If,  on  the  other  hand,  we  know  the  number  of  unit  poles  two  mag- 
nets contain  and  their  distance  apart  we  may  use  the  equation: 

mm' 


Or  we  may  find  the  field  strength  of  one  at  the  distance  d  by 
the  equation  H  =  -p  and  consider  the  action  of  this  field  on  the 
the  other  magnet  by  the  equation  F  =  mH. 

Problem  10-1.  In  how  strong  a  field  is  magnet  of  400  units 
strength  in  problem  8-1,  to  fulfill  conditions  of  the  problem? 

Problem  11-1.  Two  magnets,  one  of  800  units  pole  strength, 
the  other  of  500  units  pole  strength,  are  3  cms.  distant  from  each 
other.  In  how  strong  a  magnetic  field  may  each  be  considered 
to  lie? 

Problem  12-1.  Compute  in  three  ways  the  force  exerted  by 
the  magnets  of  problem  11  on  each  other. 

19.  Ring  Magnets.  There  are  cases  where  a  piece  of 
iron  may  be  strongly  magnetized  and  still  possess  no  poles, 
as  in  the  ring,  Fig.  12.  Since  the  lines  nowhere  come  out 
of  the  iron,  they  cannot  produce  any  poles  or  external  field. 
If,  however,  we  break  the  ring,  as  in  Fig.  13,  we  have  the 
usual  two  poles,  a  north  and  a  south,  with  an  external  field 
of  great  intensity,  though  small  in  area.  The  complete 
ring  is  used  in  some  types  of  meters  and  also  of  'transformers 


MAGNETS   AND    MAGNETISM 


17 


where  no  external  field  is  desired.  The  broken  ring  is 
the  fundamental  form  of  the  magnetic  circuit  of  generators 
and  motors  where  an  intense  external  field  is  desired  in 
limited  air  space. 


FIG.  12. — Closed  iron  magnetic  circuit.     FIG.  13. — Nearly  closed  iron  magnetic 
No  poles.  circuit. 

20.  Consequent  Poles.  Although  two  poles  is  the  least 
number  a  magnet  can  have  if  it  has  any  pole,  it  may  possess 
any  number  greater  than  two.  All  those  except  the  two 
end  poles  are  called  consequent  poles. 

Nl  and  Si  (Fig.  14),  are  consequent  poles. 


FIG.  14.     Magnet  with  consequent  poles,  Ni  and  Si. 

21.  Permanent  Magnets.  Most  of  the  electrical  measur- 
ing instruments,  such  as  ammeters,  voltmeters,  watt- 
meters, etc.,  contain  a  permanent  magnet.  The  precision 
of  the  instrument  depends  upon  the  magnet  remaining  of 
the  same  strength  from  month  to  month. 

We  have  seen  that  hard  steel  makes  the  strongest  per- 
manent magnet.  For  short  magnets  the  steel  should  be 
"glass  hard/'  that  is,  tempered  by  being  heated  to  a  red 


18  ELEMENTS   OF   ELECTRICITY 

heat  and  suddenly  immersed  in  water,  oil,  or  mercury.  A 
magnet  made  of  steel  treated  in  this  way  will  be  very  strong 
at  first,  but  even  it  will  gradually  lose  some  of  its  strength, 
especially  if  handled  roughly. 

According  to  the  "  MAGNETIC  MOLECULE  "  theory,  it  is 
easy  to  see  that  it  would  be  no  difficult  matter  to  jar  the 
molecules  from  their  magnetic  position. 

22.  Magnetic  Molecules.  Weber's  theory  of  magnetism  is 
perhaps  the  best  one  advanced  to  explain  the  different  magnetic 
phenomena. 

He  supposes  that  all  matter  is  made  up  of  small  molecules 
which  are  minute  magnets.  In  iron  and  steel,  these  little  magnets 
are  strong,  in  all  other  materials  they  are  weak.  When  a  piece 
of  material  is  not  magnetized,  these  molecules  lie  in  no  regular 
position  with  regard  to  one  another,  as  in  Fig.  15.  When  the 

FIG.  15. — Molecules  of  unmagnetized  bar. 

material  is  magnetized  the  molecules  all  lie  with  their  N  ends 
pointing  the  same  way,  as  in  Fig.  16. 

A  good  illustration  of  this  is  to  imagine  a  herd  of  cattle  crowded 
into  a  pen.  They  will  face  in  every  direction.  They  thus  fairly 
represent  the  molecules  of  an  unmagnetized  piece  of  iron.  Now 
suppose  someone  comes  to  one  end  of  the  pen  with  some  fodder. 
The  cattle  will  all  head  toward  that  end  of  the  pen.  They  then 
represent  the  position  of  the  molecules  in  a  magnetized  piece  of 
iron.  The  ease  with  which  the  molecules  can  be  turned  into 
this  position  is  partially  represented  by  the  permeability  of  the 
material;  as  to  whether  or  not  a  permanent  magnet  is  formed 
depends  on  whether  the  molecules  tend  to  retain  this  position 


FIG.  16. — Molecules  of  magnetised  bar. 


of  Fig.  16,  or  return  to  their  original  position  of  Fig.  15.  The 
molecules  of  hard  steel  tend  to  retain  this  magnetic  position, 
whereas  the  molecules  of  soft  iron  tend  to  return  to  the  original 


MAGNETS  AND    MAGNETISM  19 

position.  Thus  hard  steel  forms  a  PERMANENT  magnet,  and  iron, 
a  TEMPORARY  one. 

23.  Aging  of  Magnets.  It  has  been  found  that  by  every  jar 
that  a  magnet  receives,  it  loses  some  of  its  strength,  though  the 
amount  lost  by  each  successive  jar  becomes  less  and  less.  Thus 
all  magnets  which  are  to  be  of  unvarying  strength,  are  put  through 
a  process  of  " aging,"  which  has  the  effect  of  settling  their  strength 
at  a  definite  point.  With  reasonable  care  they  will  retain  this 
strength  indefinitely.  By  one  method  the  steel  is  steamed  for  30 
hours  or  longer,  then  magnetized  and  steamed  for  4  or  5  hours 
more.  This  treatment  demagnetizes  it  more  or  less,  of  course, 
but  the  remaining  magnetism  is  very  nearly  permanent,  especially 
if  the  steel  forms  a  nearly  closed  loop. 

It  has  also  been  found  that  thin  magnets  are  stronger  in  propor- 
tion to  their  weight  than  thick  ones,  so  a  laminated  magnet,  that 
is,  one  made  of  thin  magnets  bound  together,  is  stronger  than  one 
made  of  a  solid  piece  of  steel. 

Heat  and  rough  treatment  will  ruin  the  best  of  magnets,  hence 
the  need  of  careful  handling  of  electrical  instruments. 

The  method  of  magnetizing  iron  and  steel  will  be  discussed  in 
the  next  chapter. 


20  ELEMENTS    OF    ELECTRICITY 


SUMMARY  OF  CHAPTER  I 

MAGNETIC  FLUX  consists  of  lines  of  force  which  make 
a  complete  circuit.  Symbol,  <j>. 

THE  INTENSITY  at  any  point  in  this  magnetic  circuit,  or 
field,  is  measured  in  gausses,  which  represent  the  force  on  a 
unit  pole,  or  the  number  of  lines  per  sq.cm.  Symbol,  H. 

TOTAL  FLUX,  <£,  equals  average  intensity,  H,  times  the 
area,  A,  <£=AH. 

A  NORTH  POLE  is  the  place  where  lines  of  force  leave  a 
magnet;  a  SOUTH  POLE,  where  lines  enter.  The  poles  are 
often  considered  to  be  points.  Like  poles  repel  ;  unlike  poles 
attract  one  another. 

A  UNIT  North  pole  is  one  of  such  strength  that  if  placed 
a  centimeter  distant  from  a  like  pole  of  equal  strength,  would 
repel  it  with  force  of  one  dyne. 

47:  LINES  emerge  from  one  unit  N  pole,  and  enter  one 
unit  S  pole. 

THE  STRENGTH  of  a  magnet  may  be  stated  in  the  num- 
ber of  unit  poles  it  contains,  symbol,  m,  or  in  the  total  flux 
coming  out  of  the  N  pole  cf>.  That  is,  <£= 


FORCE  exerted  by  one  pole  on  another,  F  =  —  3-. 

FORCE,  by  magnetic  field  on  pole  placed  in  it:  F  =  mH. 

FLUX  DENSITY  equals  number  of  lines  per  sq.cm. 
Symbol,  B. 

In  air,  Flux  Density  equals  field  Intensity  equals  Magnetiz- 
ing Force;  that  is,  for  air  B  =  H. 

PERMEABILITY  is  the  relation  of  B  to  H  in  any  mate- 
rial; equals  the  flux  density  for  one  unit  of  magnetizing 
force.  Symbol  p. 


MAGNETS   AND   MAGNETISM 


21 


PROBLEMS  ON  CHAPTER  I 

13-1.  Plot  the  magnetic  field  outside  and  inside  of  a  bar  mag- 
net. 

14-1.  Show  the  flux  in  the  magnetic  circuit  of  the  two-pole 
dynamo  with  a  drum  armature,  Fig.  17.  Indicate  the  direction 
of  the  lines  of  force  by  arrow-heads. 

15-1.  Show  the  flux  in  the  magnetic  circuit  of  the  dynamo 
with  a  ring  armature,  Fig.  18.  Indicate  direction  of  the  lines  as 
in  Problem  14-1. 

16-1.  If  one  of  two  magnets  which  are  4  cms.  apart  has  a 
pole  strength  of  5  units,  what  pole  strength  must  the  other  have 


FIG.  17. 


FIG.  18. 


in  order  that  there  may  be  exerted  between  them  a  force  of  89 
dynes? 

17-1.  Draw  the  flux  in  the  magnetic  circuit  of    a    four-pole 
motor,    Fig.    19.     Indicate    the 
direction   of  the   force   lines  by 
arrow-heads. 

18-1.  A  bar  of  steel  100  cms. 
long,  4  cms.  wide,  2  cms.  thick  is 
magnetized  so  that  it  has  approx- 
imately 800  units  pole  strength 
per  sq.cm.  of  cross-section. 

(a)  What  is  its  pole  strength, 

m?  ^ 

(b)  Compute  total  flux,  <£,  pass- 
ing out  of  the  AT  pole. 

19-1.  Calculate  force  in  dynes  with  which  the  pole  in  Problem 
18-1  would  attract  the  S  pole  of  a  similar  magnet  of  equal 
strength  if  placed  20  cms.  from  it. 


FIG.  19. 


22  ELEMENTS  OF  ELECTRICITY 

20-1.  One  of  the  pole  faces  of  a  dynamo  has  an  area  of  20X30 
cms.,  and  1,800,000  lines  pass  from   the   face   into   the    smooth 

core  of  armature.      What    is   the 

intensity  of  the  magnetic  field  in 
the  air  gap  between  the  pole  face 
and  the  armature  core? 

21-1.  A  magnet  of  5  units  pole 
strength  is  in  a  field  of  16  gausses 


FIG.  20.  intensity.     How  much  is  the  force 

on  the  pole?  f 

22-1.  Draw  the  magnetic  field  of  two  magnets  placed  side  by 
side  as  in  Fig.  20. 

23-1.  Two  magnets  are  5  cms.  apart. 

w  =  280  unit  poles. 

w'  =  500  unit  poles. 
(a)  Find  force  between  them. 
(6)  In  how  strong  a  field  is  ml 
In  how  strong  a  field  is  m'  ? 

24-1.  A  magnet  of  120  units  pole  strength  is  in  a  field  of  400 
gausses. 

(a)  Force  exerted  on  magnet? 

(6)  How  far  from  another  magnet  of  1000  units  would  this  first 
magnet  have  to  be  placed  to  be  in  a  field  equal  to  400  gausses? 

(c)  What  would  be  the  force  between  the  magnets  in  (6)  ? 

25-1.  A  magnetic  pole  having  a  strength  of  1800  lines  is 
placed  in  a  field  of  2000  gausses.  What  is  the  action  on  the 
magnet? 

26-1.  A  magnetic  pole  having  90,000  lines  is  placed  in  a  mag- 
netic field  where  there  is  an  average  of  1000  lines  to  every  50 
sq.cms.  What  force  (in  pounds)  is  exerted  on  the  magnet? 

27-1.  A  magnet  has  a  pole  strength  of  500  unit  poles.  How 
many  lines  emerge  from  the  N  pole? 

28-1.  What  is  the  field  intensity,  4  cms.  from  pole  of  magnet 
in  Problem  27-1? 

29-1.  If  a  piece  of  iron,  /*  =  800,  were  placed  at  point  in 
field  indicated  in  Problem  28-1,  what  would  be  the  flux  density 
in  the  iron?  Assume  magnetizing  force  to  remain  the  same  as 
in  Problem  28-1. 

30-1.  If  iron  in  Problem  29-1  has  a  cross-section  area  of  6 


MAGNETS  AND   MAGNETISM  23 

sq.cms.  (at  right   angles   to  lines)  how  many  unit   poles  are   in- 
duced in  it? 

31-1.  With  what  force  will  iron  in  Problem  29-1  be  attracted 
to  magnet? 

32-1.  The  intensity  at  a  certain  point  in  the  magnetic  field  of  a 
magnet  was  30  gausses.  A  piece  of  soft  iron  placed  in  that  part 
of  the  field  was  magnetized  to  a  flux  density  of  48,000  gausses. 
What  was  the  permeability  of  the  iron?  Assume  H  to  remain 
constant. 

33-1.  If  the  magnetic  field  of  Problem  32  was  due  to  a  mag- 
net of  4800  unit  poles,  how  far  from  one  of  the  poles  is  the 
point  mentioned? 

34-1.  What  would  be  the  flux  density  in  the  iron  of  Problem 
32  and  33  if  it  were  moved  1  centimeter  nearer  the  pole?  Assume 
(/*)  to  remain  constant. 

35-1.  A  piece  of  wrought  iron,  the  permeability  of  which  is 
1600,  is  placed  in  the  magnetic  field  of  a  bar  magnet  of 
2500  unit  poles.  It  becomes  magnetized  with  a  flux  density  of 
50,000  gausses,  (a)  What  is  the  field  strength  at  the  point  where 
the  iron  is  placed?  (6)  How  far  is  this  point  from  the  pole  of  the 
magnet? 

36-1.  The  area  of  the  pole  face  A  of  generator  in  Fig.  149  is 
168  sq.in.  (B)  for  the  iron  equals  12,000  gausses.  How  many  lines 
emerge  from  the  pole  face? 


CHAPTER  II 


ELECTROMAGNETS 

Magnetic  Field  about  a  Straight  Wire — Thumb  Rule — Resultant  of" 
Circular  and  Parallel  Fields — Moving  Force  in  Voltmeter,  Amme- 
ter, Motor,  etc. — Field  about  a  Coil — Thumb  Rule  for  Coil — 
— Electromagnets— Magnetic  Hoists— Telegraph — Generator  and 
Motor  Fields — Sucking  Coils — Non-inductive  Coils. 

WHEREVER  there  is  an  electric  current,  there  is  also  a 
MAGNETIC  current.  Another  way  of  stating  the  same  truth 
is:  Electricity  in  motion  always  produces  a  magnetic  field. 
This  magnetic  field  is  always  at  right  angles  to  the  electric 
current  which  accompanies  it. 

24.  Field  about  a  Straight  Wire.  When  the  wire  carry- 
ing the  electric  current  is  straight,  the  magnetic  field  about 
the  wire  is  circular,  and  a  N  pole  placed  near  the  wire  would 


FIG.  21.     Magnetic  field  about  a  straight  wire  carrying  a  current  of  electricity. 
Field  is  circular,  not  spiral. 

whirl  around  and  around  the  wire  in  a  circle.  If  the  cur- 
rent in  the  wire  is  reversed,  the  N  pole  would  still  whirl 
around  the  wire  but  in  the  opposite  direction,  showing  that 
the  field  was  the  same  shape  but  opposite  in  direction. 
Fig.  21  shows  this  circular  field  about  a  straight  wire. 

24 


ELECTROMAGNETS 


25 


If  now,  we  look  along  the  wire  in  the  direction  in  which 
the  current  is  flowing,  the  magnetic  field  is  whirling  around 
the  wire  in  the  direction  we  would  turn  down  a  right-hand 
screw.  Notice  in  particular  that  these  whirls  are  not 
spirals  but  are  circles.  Fig.  22  shows  a  cross-section  of 
the  wire  and  magnetic  field,  and  represents  the  way  the 
field  would  appear  if  we  looked  at  the  end  of  the  wire  with 
the  current  going  away  from  us.  In  Fig.  23  the  current 
is  reversed.  Notice  that  the  field  is  also  reversed  in 
direction. 


FIG.  22. — Field  about  a  straight 
wire;   end  view. 


Fio.  23. — Field  about  a  wire; 
end  view;  current  reversed. 


The  best  way  for  finding  the  direction  of  the  magnetic 
field  about  a  wire  carrying  an  electric  current  is  by  the  thumb 
rule. 

26.  Thumb  Rule  (for  straight  wire).  If  we  grasp  the 
wire  with  our  right  hand,  so  that  the  THUMB  points  in  the 
direction  of  the  current,  the  FINGERS  will  point  in  the  direc- 
tion of  the  magnetic  field. 

Similarly,  if  we  know  the  direction  of  the  magnetic  field 
we  can  find  the  direction  of  the  current.  For  if  we  place 
the  fingers  in  the  direction  of  the  lines  of  force,  the  thumb 
will  then  point  in  the  direction  of  the  current. 

Figs.  24  and  25  show  this  circular  field  about  a  wire 
carrying  a  current  taken  by  means  of  iron  filings. 

26.  Circular  Field  Combined  with  Parallel  Field.  It 
will  be  remembered  that  the  field  between  two  unlike  poles 
as  in  Fig.  4  (which  see)  is  nearly  of  uniform  density. 


26 


ELEMENTS   OF  ELECTRICITY 


If  we  now  place  a  wire  in  this  field  perpendicular  to  the 
lines  of  force  and  send  an  electric  current  through  the  wire, 
this  current  sets  up  a  circular  field  about  the  wire  as  in  Fig. 
25.  But  as  this  circular  field  is  placed  in  the  uniform  field 


Direction  of 

— >• 
Earth's  Field 


FIG.  24. — Field  about  a  wire  shown  by  iron  filings. 

of  Fig.  4,  the  result  is  the  combination  field  in  Fig.  26.     The 
lines  are  very  much  more  crowded  on  the  upper  side  of  the 


FIG.  25. — Field  about  a  wire  shown 
by  iron  filings. 


FIG.  26. — Combination  of  circular  and  paral- 
lel magnetic  fields;  shown  by  iron  filings. 


wire  than  on  the  lower  side.  The  lateral  pressure  (crowd- 
ing effect)  of  the  lines  on  the  upper  side  is  therefore  greater 
than  the  lateral  pressure  below  the  wire.  Therefore  the 
wire  will  be  forced  down. 

This  is  the  ^fundamental  principle  of  electric  motors, 
Weston  voltmeters  and  ammeters,  D'Arsonval  galvanom- 
eters, etc.  A  large  torque  is  obtained  by  having  a  very 
strong  field  with  many  wires  in  it,  carrying  heavy  currents. 


ELECTROMAGNETS 


27 


The  sum  of  the  pressures  on  all  the  wires  can  thus  be  made 
to  amount  to  a  very  great  force. 

27.  Explanation  of  Shape  of  Resultant  Field.  The 
cause  of  the  lines  being  more  crowded  on  one  side  of  the 
wire  is  due  to  the  circular  shape  of  the  field,  which  makes 
the  direction  of  the  force  lines  on  one  side  of  the  wire  to  be 
exactly  opposite  to  the  direction  of  those  lines  on  the  other 
side.  In  Fig.  27,  notice  that  the  lines  above  the  wire  flow  to 
the  right,  below  to  the  left.  When  this  wire  is  placed  in  uni- 
form field  flowing  to  the  right,  as  in  Fig.  28,  the  lines  above 


FIG.  27.— Field  about  a  straight 
wire. 


FIG.  28. — Wire  carrying  current  placed  in 
parallel  magnetic  field. 


the  wire,  flowing  to  the  right,  join  those  of  the  field  flowing 
to  the  right  and  thus  strengthen  the  field  above.  Those 
below  the  wire,  flowing  to  the  left,  neutralize  some  of  those 
of  the  field  flowing  to  the  right,  and  weaken  the  field  below 
the  wire.  Thus  a  strong  field  above  the  wire  and  a  weak 
field  below,  resulting  in  a  force  urging  the  wire  down  as 
previously  seen. 

28.  D.  C.  Voltmeters,  Ammeters,  etc.  The  Weston 
D.  C.  ammeters  and  voltmeters  illustrate  the  practical 
application  of  this  force  existing  between  a  parallel  field 


28 


ELEMENTS  OF  ELECTRICITY 


and  a  wire  carrying  a  current.     See  Figs.  29  and  30,  also 
Figs.  290  and  291. 

A  coil  AB  is  carefully  set  on  jeweled  bearings  between 
the  poles  of  a  permanent  aged  magnet  N  and  S  as  described 
in  Chapter  I.  Watchsp rings  W  hold  the  coil  in  place,  so 
that  the  pointer  is  held  at  zero  on  the  scale,  when  no  current 
is  flowing  through  the  coil.  Suppose  a  current  is  led  into 
the  moving  coil  so  that  it  goes  down  along  the  side  B  and 
up  the  side  A.  On  a  top  view,  Fig.  30,  the  current  would 
go  in  at  B  and  out  at  A .  The  clockwise  circular  field  around 
B  would  strengthen  the  field  of  the  permanent  magnet 


FIG.  29. — Moving  coil  of  Weston 
D.C.  meter. 


FIG.  30. — Construction  of  Weston 
D.C.  meter. 


NSt  above  the  wires  B  and  weaken  it  below.  This  would 
urge  the  side  B  downward.  In  the  same  way  the  counter- 
clockwise field  about  A  would  strengthen  the  magnet's 
field  below  A  and  weaken  it  above.  Thus  A  would  be 
urged  upward. 

These  two  actions  would  cause  the  coil  to  turn  against 
the  tension  of  the  springs  W.  The  stronger  the  current 
flowing  through  the  coil  the  stronger  the  combination  fields 
causing  the  coil  to  turn.  The  pointer  would  then  indicate 
the  current  as  it  moved  over  a  scale  graduated  in  amperes. 
The  instrument  is  then  an  ammeter.  As  the  current  in  the 


ELECTROMAGNETS  29 

coil  is  proportional  to  the  voltage  across  its  terminals,  the 
amount  the  coil  turns  must  also  be  proportional  to  the 
voltage.  The  scale  accordingly  might  be  graduated  to 
read  volts.  It  would  then  be  a  voltmeter.  These  instru- 
ments are  described  fully  in  a  later  chapter,  Figs.  280-9, 
show  more  complete  details  of  the  various  types.  For 
the  similar  action  between  the  armature  and  field  of  a  motor 
see  Fig.  165  and  description  relating  to  it. 

29.  Field  about  a  Coil  Carrying  a  Current.  If  a  wire  is 
made  into  a  loop  as  in  Fig.  31,  we  find,  by  the  THUMB  rule, 
that  the  lines  of  force,  which  everywhere  whirl  around  the 
wire,  all  enter  the  same  face  of  the  loop  and  all  come  out 
of  the  other  face. 


FIG.  31. — Field  about  a  single  loop  carrying  a  current. 

If  we  now  place  several  loops  together  into  a  loose  coil 
as  in  Fig.  32,  most  of  the  lines  will  thread  the  whole  coil. 
If  we  make  a  close  coil,  practically  all  the  lines  will  thread 
the  whole  coil,  and  return  outside  the  coil  to  the  other  end. 

The  reason  that  practically  no  lines  of  force  encircle  the  separate 
loops  of  a  closely  wound  coil,  but  all  thread  the  entire  coil,  is  ex- 
plained by  referring  to  Fig.  33.  This  drawing  represents  an  en- 
larged longitudinal  section  of  the  coil  in  Fig.  32.  The  current 
entering  the  ends  of  half  loop  at  A,  B,  and  C,  comes  out  again 


30 


ELEMENTS   OF  ELECTRICITY 


at  D,  E,  and  F.  If  the  wire  ends  A  and  B  were  pushed  nearer 
one  another,  the  field  on  the  right  side  of  A  would  be  in  the  opposite 
direction  and  would  neutralize  the  field  on  the  left  side  of  B. 


FIG.  32.  —  Field  within  and  without  a  loose  coil  carrying  a  current. 

The  space  between  the  wires  A  and  B  would  thus  be  neutral  or 
free  of  lines  of  force.  The  lines  now  would  be  compelled  to  con- 
tinue on  through  the  whole  length  of  the  coil,  and  would  not  slip 


FIG.  33. — Longitudinal  cross-section 
coil  in  Fig.  32. 


spaces  between  the  loops 
and  encircle  each  wire  with  a  sep- 
arate field. 

We  thus  have  the  same 
shaped  field  as  in  and  about  a 
bar  magnet;  one  end  being  a 
N  pole,  since  all  the  lines  come 
out  of  it,  and  the  other  end 
a  S  pole,  since  all  the  lines 
enter  it.  In  order  to  find  the 
polarity  of  such  a  coil,  it  is  best 

to  use  a  modified  form  of  the  Thumb  Rule  for  a  straight 

wire. 

30.  Thumb    Rule     (for  coil).     Grasp  the  coil  with  the 
right  hand  so  that  the  FINGERS  point  in  the  direction  of  the 
current  in  the  coil,  and  the  THUMB  will  point  to  the  N  pole. 
N.  B.  —  This  is  the  thumb  rule  for  the  straight  wire  reversed. 

31.  Coil    Equivalent    to    a    Bar    Magnet.     A  coil  with  a 
current  flowing  through  it,  is  then  the  equivalent  of  a  bar 
magnet  with  two  poles.     It  obeys  the  laws  of  a  bar  magnet 
as  stated  in  the  first  chapter.     That  is,   unlike  poles  of  two 
coils  attract  each  other  and  like  poles  repel;    if  the  coil  is 
free  to  turn  and  is  placed  in  a  field,  it  will  tend  to  take  such 


ELECTROMAGNETS  31 

a  position  that  the  lines  inside  the  coil  are  parallel  to  the 
lines  in  the  field. 

32.  Electromagnets.     Very  strong  magnets  are  made  by 
inserting  a  piece  of  iron  or  steel,  called  a  core,  in  the  coil. 
The  permeability  of  iron  and  steel  is  so  much  greater  than 
air,  as  explained  in  Chapter  I,  that  the  same  current  in  the 
same  coil  sets  up  thousands  of  times  as  many  lines  in  the 
iron  core  as  it  would  in  the  air  alone. 

33.  Magnetic    Hoists.      The  powerful  poles  of  magnetic 
hoists  are  built  in  this  way:     Coils  of  wire  are  wound  around 
a  number  of  cores  of  soft  iron.     As  long  as  there  is  no 
electric  current  going  through  the  coils  the  iron  is  not  a 
magnet.     The  face  of  the  iron  cores  is  brought  in  contact 
with  pieces  of  iron  or  steel  castings,  etc.,  and  the  current 
turned  on.     The  iron  cores  now  become  such  strong  magnets 
that  each   square   inch  of    their  ends  will   lift  from  100  to 
200  Ibs.  of  iron. 

To  release  the  load  of  iron  it  is  necessary  to  turn  off  the 
electric  current,  or  reverse  the  direction  of  it. 

Figs.  34  and  35  (Plate'  I) ,  show  the  details  of  the  Browning 
Lifting  Magnet. 

34.  The  Telegraph.     The  sounder  of  the  telegraph  sys- 
tem consists  of  two  coils  with  soft  iron  cores  which  attract 
a  small  piece  of  soft  iron  to  themselves,  when  a  current  is 
sent  through  them,  and  release  it  as  the  current  is  turned 
off,  allowing  it  to  be  pulled  away  by  the  spring. 

35.  Generator  and  Motor    Fields.      But  the  most    im- 
portant use  of  electromagnets  is  in  generators  and  motors, 
where  they  are  used  to  create  the  intense  magnetic  fields 
necessary  in  the  economical  generation  of  electric  power. 
In  Fig.  36,  which  represents  a  2-pole  motor,  the  coils  A  A 
are  wound  on  soft  iron  cores  in  such  a  way  as  to  make  one 
pole  face  a  north  pole  and  the  other  a  south  pole.     The  soft 
iron  yoke  Y  joins  the  two  cores  together  at  one  end,  so  that 
the  magnetic   circuit  is  composed  entirely  of  iron  except 
the  space  G  where  the  armature  goes,  a  large  part  of  which 


32 


ELEMENTS  OF  ELECTRICITY 


is  also  made  of  soft  iron.     The  magnetic   circuit  is  then 
a  nearly  closed  iron  circuit  in  the  direction  indicated. 
Fig.  37  represents  a  4-pole  generator  field  and  field  coils. 


FIG.  36. — Magnetic  field  of  2-pole 
motor. 


FIG.  37. — Magnetic  field  of  a  4-pole 
motor. 


36.  "Sucking"  Coils.  Just  as  a  bar  magnet  will  attract 
a  piece  of  iron,  so  a  coil,  since  it  has  a  field  like  a  bar  magnet, 
will  attract  a  piece  of  iron.  When  the  plunger  of  soft  iron, 
A,  in  Fig.-  38,  is  placed  in  the  field 
of  coil  B  it  becomes  magnetized  as 
marked,  the  N  pole  being  nearest  the 
S  pole  of  the  coil.  Thus  it  is  attracted 
into  the  coil,  being  drawn  up  until  the 
centers  of  coil  and  plunger  coincide. 
The  strongest  attraction  exists  when 
the  center  of  the  iron  plunger  nearly 
coincides  with  the  center  of  the  coil. 
Extensive  use  is  made  commercially 
of  this  principle.  It  is  the  device  used 
to  regulate  the  arc  of  the  arc  lamp, 
Fig.  39.  (For  a  full  description,  see 
Chapter  XI.)  Also  to  operate  a  circuit 
breaker,  Fig.  40.  In  the  latter  of  these  the  plunger  is  only 
sucked  up,  when  the  current  in  the  coil  becomes  excessive. 
Then  it  automatically  opens  the  switch,  shutting  off  the  power. 


FIG.  38. — Sucking  coil  at 
tracting  iron  plunger. 


ELEC  TROMAG  NE  TS 


33 


FIG.  36. — G.  E.  arc  lamp.  Lower 
coils  suck  up  plunger  of  soft 
iron. 


FIG.  40. — G.  E.  circuit  breaker. 


37.  Non-inductive  Coils.  Since  the  direction  of  the 
magnetic  field  about  a  wire  depends  on  the  direction  of  the 
electric  current,  it  is  seen  that  the  fields  of  two  currents 
flowing  in  opposite  directions  must  oppose  each  other,  and 
will  neutralize  each  other  if  the  wires  are  near  enough 
together.  Use  is  made  of  this  fact  whenever  it  is  desired 
to  have  a  current  with  a  very  weak  magnetic  field.  Two 
wires  are  wound  into  a  single  coil  so  that  the  current 
in  one  wire  flows  in  one  direction,  and  the  current  in  the 
other  wire  in  the  other  direction,  around  the  coil.  Such 
a  coil  has  no  perceptible  magnetic  field,  and  is  said  to  be 
NON-INDUCTIVELY  wound.  Resistance  coils  used  in  the  meas- 
urement of  resistance  are  wound  in  this  way,  as  are  almost 
all  resistance  coils  in  commercial  ammeters  and  voltmeters. 
Fig.  116  shows  such  a  coil  taken  from  a  Wheatstone  bridge. 


34  ELEMENTS  OF  ELECTRICITY 


SUMMARY  OF  CHAPTER  II 

There  is  always  a  magnetic  field  about  a  wire  carrying  a 
current  of  electricity. 

THUMB  RULE  FOR  STRAIGHT  WIRE.  Grasp  the  wire 
with  the  right  hand  so  that  the  thumb  points  in  the  direc- 
tion of  the  current,  the  fingers  then  will  point  in  the  direction 
of  the  magnetic  field  around  the  wire.  The  magnetic  field 
about  a  wire  when  combined  with  a  uniform  field,  causes 
a  crowding  of  the  lines  on  one  side  of  the  wire,  and  a  thinning 
out  of  the  lines  on  the  other  side.  The  wire  tends  to  move 
toward  the  thinner  part  of  the  field.  This  is  the  moving 
force  of  an  electric  motor. 

An  electric  current  flowing  in  a  coil  of  wire  makes  an  electro- 
magnet of  the  coil,  one  face  becoming  a  North  pole  and  the 
other  a  South  pole. 

THUMB  RULE  FOR  A  COIL.  Grasp  the  coil  so  that  the 
Fingers  point  in  the  direction  of  current  around  coil,  and 
Thumb  points  to  the  North  pole.  This  is  another  form  of 
the  rule  for  straight  wire. 

A  coil  made  thus  and  free  to  turn,  is  placed  between  the 
poles  of  a  permanent  magnet.  On  sending  a  current  through 
it,  the  coil  turns  proportionately  to  the  current.  Tljis  is  the 
principle  of  many  Ammeters  and  Voltmeters. 

The  powerful  magnets  composing  the  magnetic  field 
of  generators  and  motors  are  made  by  coiling  insulated 
wire  around  soft  iron  cores. 


ELECTROMAGNETS 


35 


PROBLEMS  ON  CHAPTER  II 

1-2.  Draw  the  magnetic  field  about  the  wire  A,  Fig.  41,  when 
the  current  is  flowing  as  indicated. 


FIG.  41. 

2-2.  If  A,  Fig.  42,  represents  cross-section  of  wire  with  the 
current  coming  out,  and  the  poles  of  the  motor  are  as  marked, 
draw  resulting  field  between  AT  and  S  pole.  In  what  direction 
will  wire  A  tend  to  move? 

3-2.  A  and  B,  Fig.  43,  represent  the  cross-section  of  a  loop 
of  wire  on  an  armature  in  the  magnetic  field  of  a  motor.  If 
the  current  is  as  marked  jn  A  and  B  and  poles  of  motor  are  as 
marked,  in  which  direction  will  armature  rotate? 


N 


FIG.  42. 


FIG.  43. 


4-2.  Draw  magnetic  internal  and  external  field  for   iron    core 
with  electric  current  flowing  around  it  as  indicated  in  Fig.  44. 


FIG.  44. 


D 


\ 


FIG.  45. 


5-2.  Draw  field  between  coils  A  and  B,  in  Fig.  45.  State 
whether  attraction  or  repulsion  exists  between  them. 

6-2.  Show  windings  on  cores  A  and  B,  with  direction  of  elec- 
tric current  indicated  and  draw  flux  for  magnetic  circuit  with 
direction  of  flux  indicated  in  Fig.  46. 


36 


ELEMENTS   OF  ELECTRICITY 


7-2.  Draw  6-pole  dynamo  field,  showing  circuits  and  direction 
of  flux,   and  electric  current. 


FIG.  46. 


FIG.  47.     Consequent  pole  motor  frame. 

8-2.  Show  direction  of  current  in  windings  on  A  and  B  to  pro- 
duce Poles  N  and  S,  as  marked  in  Fig.  47.  Show  magnetic  flux 
as  usual. 


FIG.  48. 


9-2.  Show  current  in  windings  on  A  to  produce  N  and  S  poles 
;  marked  in  Fig.  48.     Indicate  magnetic  flux  as  usual. 


CHAPTER  III 
OHM'S  LAW 

The  Electric  Current — Units;  Ampere  (current);  Volt  (pressure); 
Ohm  (resistance) — Analogy  of  Electrical  to  Hydraulic  Units — 
Diagrams  of  Electric  Circuits — Table  of  Convenient  Symbols — 
Voltage,  the  Essential  Factor— Potential — Electrical  Potential 
— Brush  Potential — Fall  of  Potential  along  a  Uniform  Wire — 
Ohm's  Law — Applications  of  Ohm's  Law — Series  and  Parallel 
Circuits — Voltage,  Current,  and  Resistance  Relations  in  Series 
and  Parallel  Circuits — Conductance. 

38.  The     Electric     Current.     The   nature   of   electricity 
has  not  yet  been  discovered.     It  is  certain,  however,  that 
it  is  neither  MATTER  nor  ENERGY  in  any  form,  and  therefore 
cannot  be  a  fluid.     Yet  the  flow  of  electricity  along  a  wire 
is  very  much  like  flow  of  water  through  a  pipe.     This 
analogy  to  the  flow  of  water  is  helpful  in  explaining  the 
flow  of  electricity,  and  will  be  used  to  show  the  action  of  an 
electric  current. 

39.  Ampere     (current).     A  current  of  water  in  a  pipe 
is  measured  by  the  amount  of  water  that  flows  through  the 
pipe  in  a  second:    as,  1  gal.  per  sec.,  8  gals,  per  sec.,  etc. 
In  the  same  way,  a  current  of  electricity  is  measured  by 
the  amount  of  electricity  that  flows  along  a  wire  in  a  second 
as,  1  coulomb   per  sec.,  8    coulombs  per  sec.,  etc.      The 
coulomb  of  electricity  is  merely  a  quantity  of  electricity, 
just  as  a  gallon  is  a  quantity  of  water.     Fortunately  we 
have  a  special  name  for  this  rate  of  flow  of  1  coulomb  per 
sec.,  which  is  1  AMPERE.     This  way  of  naming  the  rate  of 
flow  relieves  us  of  all  necessity  of  saying  per  second  each  time, 
as  the  second  is  part  of  the  unit,  the  ampere.    Thus  8  cou- 

37 


38  ELEMENTS  OF  ELECTRICITY 

lombs  per  sec.  is  just  8  amperes;  25  coulombs  per  sec.  is 
25  amperes,  etc. 

Inasmuch  as  we  are  usually  interested  not  in  the  amount 
of  electricity  alone,  but  in  the  amount  that  flows  in  a  second, 
the  coulomb  is  very  rarely  mentioned  except  as  a  base  to 
define  some  other  unit. 

We  have  no  general  unit  of  rate  of  flow  of  water,  so  we 
always  have  to  use  some  such  cumbersome  expression  as 
gallons  per  sec.  or  cubic  feet  per  sec. 

40.  Volt  (pressure).  The  number  of  gals  per  sec.  of 
water  flowing  through  a  pipe  depends  to  a  large  extent  on 
the  PRESSURE  under  which  it  flows.  This  water  pressure  is 
measured  in  pounds  per  square  foot.  In  the  same  way,  the 
number  of  amperes  (coulombs  per  sec.)  of  electricity  flow- 
ing along  a  wire  depends  in  part  on  the  pressure  under  which 
the  electricity  flows.  The  electrical  unit  of  PRESSURE  is 
the  VOLT.  A  volt  means  the  same  thing  in  speaking  of  a 
current  of  electricity  that  a  pound  pressure  does  in  speaking 
of  a  current  of  water.  Just  as  a  higher  pressure  is  required 
to  force  the  same  current  of  water  through  a  small  pipe 
than  through  a  large  pipe,  so  a  higher  electrical  pressure 
is  required  to  force  the  same  current  of  electricity  through 
a  small  wire  than  through  a  large  wire.  The  voltage 
(pressure)  between  two  points  in  an  electric  circuit 
is  sometimes  spoken  of  as  the  difference  in  potential, 
or  the  drop  in  potential  or  merely  the  "  drop,"  between 
those  two  points.  In  the  same  way,  the  pressure 
between  two  points  in  a  stream  of  water  (as  the  top 
and  bottom  of  a  dam),  is  often  spoken  of  as  the  difference 
in  level  or  the  drop  in  level,  or  merely  the  "  drop,"  between 
those  two  points. 

The  distinction  between  AMPERES  and  VOLTS  should 
now  be  plain.  The  amperes  represent  the  amount  of  the 
current  flowing  through  a  circuit;  the  volts  represent  the 
pressure  causing  it  to  flow. 

In  the  case  of  both  water  and  electricity  there  may  be  a 


OHM'S    LAW 


39 


great  pressure  and  yet  be  no  current.  If  the  path  of  the 
water  is  blocked  by  a  valve  being  turned  the  wrong  way, 
there  will  be  no  current,  yet  there  may  be  a  high  pressure. 
If  the  path  of  the  electricity  is  blocked  by  a  switch  being 
thrown  the  wrong  way,  there  will  be  no  current  (amperes), 
though  the  pressure  (volts)  may  be  high. 

There  is,  therefore,  something  in  addition  to  the  pressure, 
that  determines  the  amount  of  the  current,  both  of  water 
and  electricity.  This  something  is  the  resistance  of  the  pipes 
in  the  case  of  water,  and  the  resistance  of  the  wires  in  elec- 
tricity. The  greater  this  resistance,  the  less  the  current 
under  the  same  pressure. 

41.  Ohm     (resistance).     There    is    no    general    unit    to 
measure  the  resistance  which  a  pipe  offers  to   the  flow  of 
water.     The  electrical  unit  of  RESISTANCE  is  the  OHM. 

We  say  a  wire  has  ONE  OHM  resistance  when  a  pressure 
of  ONE  VOLT  forces  a  current  of  ONE  AMPERE  through  it. 
If  the  resistance  were  2  ohms,  or  twice  as  great,  the  current 
would  be  only  half  as  large:  that  is,  one-half  ampere  would 
flow. 

42.  Comparison  of  Electrical  and  Hydraulic  Units. 


UNITS 

ELECTKICTY 

WATER 

QUANTITY 

COULOMB 

GALLON 

CURRENT 

AMPERE 
1  coulomb  per 
second 

GALLON  per  min. 

PRESSURE 

VOLT 

POUNDS  per  sq.in. 

RESISTANCE 

OHM 

43.  Diagrams  of  Electric  Circuits.  In  order  that  the 
diagrams  used  later  throughout  the  text  may  give  no  trouble, 
a  few  of  the  symbols  in  common  use  are  indicated  in  this 
paragraph. 


40 


ELEMENTS   OF  ELECTRICITY 


An  electric  current  is  always  thought  of  as  flowing  from 
a  higher  to  a  lower  level.  We  mark  the  higher  level  +  , 
and  the  lower  — ,  in  order  to  denote  in  what  direction  the 
current  is  flowing.  Sometimes  arrow-heads  are  also  put 
on  the  wire.  The  current  always  flows  from  the  -f  to 
the  — .  A  given  point  is  then  +  to  all  points  below  its 
level  and  —  to  all  points  above  its  level.  In  Fig.  49 
suppose  the  current  to  flow  from  A  to  C.  If  we  are  consider- 
ing points  A  and  B,  the  point  A  would  be  +  and  the  point 
B  — .  But  if  we  are  considering  B  and  C,  B  would  then 
be  +  and  C  — . 


FIG.  49. — Diagram  of  electric  circuit 
with  battery  cell. 


Fio.  50. — Terminals  of  electric 
circuit. 


Or  consider  the  terminals  A  and  B,  in  Fig.  50.  Since  A 
is  marked  +,  it  means  that  the  current  will  flow  from  A 
and  down  to  B  when  the  two  points  are  connected  by  an 
electric  circuit. 


FIG.  51. — Diagram  of  generator  delivering 
power  to  lamps  and  motor. 


FIG.  52. — Diagram  of  electric 
circuit. 


The  terminals  on  all  D.C.  instruments  are  marked  in  this 
way,  which  is  to  indicate  that  the  +  terminal  is  always 
to  be  connected  to  the  higher  level  of  the  circuit. 

To  represent  a  battery  cell,  we  use  the  symbol   |l,  the 


OHM'S  LAW  41 

longer  line  being  the  +,  or  the  one  from  which  the  current 
flows  out  of  the  cell. 

44.  Table  of  Convenient  Symbols. 

Battery  cell, 

Generator,  D.C. 

Generator,  A.C. 

Motor 

Incandescent  lamp, 

Arc  lamp, 

Non-inductive  resistance, 

Inductive  resistance 

Switch— Single-throw, 

Switch — Double-throw, 

Galvanometer, 

Voltmeter, 

Ammeter, 

Accordingly  Fig.  51  represents  a  D.  C.  Generator  lighting 
a  bank  of  4  incandescent  lamps  (L)  and  driving  a  motor 
M.  The  current  is  flowing  from  the  generator  along  the 
top  of  the  circuit  to  lamps  and  motor. 

Sometimes  merely  the  terminals  to  the  source  of  the 
power  are  shown  as  A  and  B  in  Fig.  52. 

A  =power  terminal;         R  ^resistance. 

B  =power  terminal;          L  =  incandescent  lamp. 

The  current  comes  from  A,  goes  through  R,  then  L  and 
finally  leaves  at  B. 

45.  Pressure,  the  Essential  Factor.      When  we  desire  to 
have  water  piped  to  a  house,  the  one  thing  above  all  others 
which  we  must  have,  in  order  to  secure  any  flow  at  all, 
is  PRESSURE.     The  water  company  agrees  to  furnish  this 
water  pressure.     The  flow  of  water  into  your  house  will 
then  depend  entirely  upon  your  wishes.     You  have  the 


42  ELEMENTS  OF   ELECTRICITY 

desired  pressure  always  at  hand  and  can  allow  it  to  force 
a  great  quantity  of  water  through  your  pipe  or  only  a  small 
amount.  You  pay  for  the  amount  of  water  you  allow 
this  pressure  to  force  through  your  pipes.  If  it  had  been 
necessary  to  maintain  twice  as  great  a  pressure  in  order  to 
force  this  same  amount  of  water  through  your  pipes,  the 
company  might  charge  you  twice  as  much  for  the  water 
used.  In  other  words,  they  would  charge  you  for  the 
work  done  in  pumping  so  much  water  into  your  house. 
At  a  given  pressure,  this  amount  of  work  done  would  be 
proportional  to  the  amount  of  water  which  you  allowed 
to  flow  into  your  house. 

In  the  same  way,  if  you  desire  to  have  electricity  brought 
to  your  house,  the  one  thing  above  all  others  that  is  neces- 
sary, in  order  to  get  a  flow  of  electricity,  is  that  an  electric 
pressure  be  maintained.  The  electric  company  agrees  to 
maintain  this  electric  pressure.  The  flow  of  the  current 
into  your  house  depends  entirely  upon  your  wishes.  The 
pressure  is  always  at  hand.  You  may  allow  it  to  cause  a 
large  current  to  flow  to  the  house  or  only  a  small  current. 
You  pay  for  the  amount  of  electricity  you  use,  at  that 
pressure.  If  the  electric  company  maintained  twice  as 
much  pressure  and  you  used  the  same  amount  of  electricity, 
you  would  have  to  pay  twice  as  much  for  this  quantity. 
In  other  words,  they  would  charge  you  for  the  work  done 
in  forcing  so  much  electricity  into  your  house.  At  a  given 
pressure,  the  amount  of  work  done  would  be  proportional 
to  the  amount  of  electricity  which  you  allowed  to  flow  into 
your  house. 

The  starting  point  of  all  hydraulic  and  electric  proposi- 
tions is  thus  the  PRESSURE.  There  can  be  no  current  either 
of  water  or  electricity  without  pressure.  The  amount  of 
work  done  in  a  staged  time  by  a  given  current  depends 
entirely  upon  PRESSURE.  It  is  necessary,  then,  to  study 
this  feature  of  pressure  a  little  more  in  detail. 

46.  Potential.     In  order  to  exert  a  pressure  tending  to 


OHM'S  LAW 


43 


A 

R 
0 



R 
O 
B 

1 

T 

cause  water  to  flow  toward  a  given  point,  it  is  merely  nec- 
essary to  raise  the  water  to  a  level  above  that  point.  That 
is,  we  give  it  potential  energy  with  regard  to  that  point, 
by  the  raising  of  it  to  a  point  which  has  a  higher  level  or 
higher  POTENTIAL  than  that  point.  Thus  if  we  wish  to 
cause  water  to  flow  out  of  tap  A  (Fig.  53),  we  may  connect 
it  to  a  tank  T.  As  long  as  the  water  in  the  tank  is  kept 
above  the  level  A BB,  the  water  will  flow  from  T  to  A, 
because  it  is  at  a  higher  level  or  potential  in  T  than  in  A. 
This  difference  in  potential  then,  between  two  points  is  what 
causes  a  pressure  between  these  two  points.  The  greater 
this  difference  of  poten- 
tial the  greater  the  pres- 
sure. When  the  difference 
of  potential  between  the 
level  of  the  water  in  T 
and  A  is  only  OB,  the 
pressure  is  less  than  when 
the  difference  of  poten- 
tial is  RB.  The  differ- 
ence of  potential  is  then 
a  measure  of  the  pres- 
sure. It  is  a  matter  of 
common  experience  that 
there  is  always  a  tendency  for  the  water  to  flow  toward  the 
lower  potential. 

47.  Electrical  Potential.  Similarly,  if  we  wish  to  cause 
a  current  of  electricity  to  flow  from  one  point  to  another, 
we  have  merely  to  raise  the  potential  of  the  first  point 
above  that  of  the  second.  Then  a  pressure  is  set  up  propor- 
tional to  their  difference  in  potential.  This  difference  of 
potential  tends  to  send  a  current  from  the  higher  potential 
to  the  lower,  as  in  the  case  of  water. 

Electricity  always  flows  from  a  high  potential  point  to  a 
lower.  A  battery  or  generator  may  then  be  thought  of,  as 
being  merely  a  pump  which  keeps  one  end  of  the  line  at  a 


FIG.  53. 


•Water  pressure  at  A  depends  upon 
water  level  in  tank  T. 


44 


ELEMENTS  Of   ELECTRICITY 


higher  potential  than  the  other,  thus  setting  up  a  pressure 
between  these  two  points.  Accordingly,  the  electric-power 
company  runs  two  wires  to  your  house  and  simply  agrees 
to  keep  the  difference  in  the  potential  between  them  up  to 
a  certain  value;  providing,  of  course,  you  do  not  allow  so 
much  electricity  to  flow  that  the  generator  is  unable  to 
pump  fast  enough  to  keep  up  this  difference  of  potential. 
In  such  a  case  it  would  be  just  as  if  the  water  in  Fig.  53 
were  allowed  to  flow  out  of  A  so  fast  that  the  pump  M 
could  not  keep  the  level  in  the  tank  above  the  line  ABB. 

In  referring  to  Fig.  49,  we  have  said  that  the  point  A  was 
at  a  higher  level  than  B,  and  B  higher  than  C.     We  may  now 


A 

B 

C 

D 

E 

F         G 

H         1 

1 
0       10 

1 
20 

1 
30 

1 

40 

1 
50 

1         1 
60        70 

,  • 

1         I 
80         90       1( 

FIG.  54 — Drop  in  potential  along  a  uniform  wire. 

say  that  A  is  at  a  higher  potential  than  B,  and  B  than  C. 
A  current  therefore  flows  from  A  to  B  and  from  B  to  C,  due 
to  these  differences  in  potential.  The  battery  acts  as  a 
pump  to  keep  the  left-hand  side  continually  at  a  higher 
potential  than  the  right-hand  side.  The  difference  of 
potential  between  A  and  B  is  sometimes  spoken  of  as  the 
fall  of  potential  from  A  to  B,  or  as  the  drop  in  potential 
from  A  to  B.  The  same  applies  to  B  and  C,  or  any  other 
points.  The  difference  in  potential  between  two  points 
in  an  electric  circuit  is  called  the  drop  in  potential  for  that 
part  of  the  circuit  contained  between  those  two  points,  and  is 
the  cause  of  any  current  flowing  between  these  two  points. 

In  Fig.  52  the  terminal  A  is  attached  to  the  high-poten- 
tial side  of  the  generator,  since  it  is  marked  +  and  B  is  at- 
tached to  the  low-potential  side  of  the  generator,  since  it  is 


OHM'S   LAW  45 

marked  — .  The  current  therefore  flows  from  A  through  R, 
then  through  L  to  B,  and  back  to  the  low-potential  side  of 
the  generator.  The  generator  must  continually  raise  elec- 
tricity up  to  the  high-potential  side,  to  make  up  for  that 
which  flows  away  from  that  side  to  the  low-potential  side. 
If  no  electricity  is  allowed  to  flow  away  from  A  to  B, 
then  the  generator  has  to  raise  no  more  electricity  up  to  the 
potential  of  A.  It  merely  has  to  keep  the  pressure  up. 

48.  Fall  of  Potential  along  a  Uniform  Wire.  In  Fig. 
54  a  uniform  wire  OABC  .  .  . ,  100  ft.  long,  has  a  current 
sent  through  it  from  0  to  H  by  means  of  a  generator  X. 
The  point  0  must  be  a  higher  potential  than  H,  since  the 
current  is  flowing  from  0  to  H.  ' 

Let  us  mark  this  wire  into  10-ft.  lengths  and  letter  the 
points  A,  B,  C,  etc.,  beginning  10  ft.  from  0.  The  point 
0  will  now  be  at  a  higher  potential  than  A,  A  than  B,  B 
than  C,  etc.  If  we  now  measure  by  means  of  some  instru- 
ment the  difference  in  potential  between  any  two  of  these 
points  10  ft.  apart,  we  find  the  difference  of  potential  is 
the  same  between  any  two.  This  shows  that  the  drop  of 
potential  across  equal  distances  along  a  uniform  wire  is 
the  same. 

The  instrument  used  to  measure  drop  or  difference  of 
electrical  potential  is  called  VOLTMETER.  It  indicates  this 
potential  difference  in  VOLTS. 

If  we  now  place  the  voltmeter  across  OB  (twice  the 
length  of  OA)  and  measure  the  drop  between  the  points 
0  and  B,  we  find  that  the  drop  is  twice  that  of  OA.  In  the 
same  way  the  drop  across  OC  (three  times  the  length  of 
OA)  is  three  times  the  drop  across  OA.  This  holds  true 
for  any  multiple  of  OA,  and  proves  that  the  drop  across 
parts  of  a  uniform  wire  carrying  an  electric  current  is  pro- 
portional to  the  length  of  these  parts.  Now  since  the 
resistance  of  a  wire  is  also  proportional  to  its  length,  we  may 
say: 

The  drop  in  potential  across  any  part  of  an  electric  circuit 


46  ELEMENTS  OF  ELECTRICITY 

is  proportional  to  the  resistance  of  that  part,  if  the  same  current 
is  flowing  through  each  part. 

Example.  The  voltage  across  6  ohms  resistance  is  20  volts. 
What  will  be  the  drop  across  30  ohms,  if  the  same  current  is  flow- 
ing through  each  resistance? 

Let  F  =  drop  across  the  30  ohms. 

Since  the  drops  are  proportional  to  the  resistances,  the  current 
being  the  same, 

20=~6~ 
V  =  100  volts. 

Problem  1-3.  If  the  drop  along  40  ft.  of  the  line  wire  in  a  cer- 
tain circuit  is  .002  volt,  what  is  the  drop  per  mile? 

Problem  2-3.  The  drop  across  20  ft.  of  wire  is  .06  volt.  The 
drop  across  an  unknown  length  is  14.4  volts.  How  great  is  un- 
known length? 

Problem  3-3.  Two  resistances  are  connected  in  such  a  way  that 
the  same  current  flows  through  each.  The  resistance  of  one  is 
.1  ohm,  the  other  is  unknown.  The  voltage  across  the  known  is 
.48  volt,  across,  the  unknown  is  4.62  volts.  What  is  the  value 
of  the  unknown  resistance? 

49.  Ohm's  Law.  The  relation  between  current  (amperes) , 
pressure  (volts),  and  resistance  (ohms),  is  stated  by  the 
famous  Ohm's  law.  This  law  is  the  beginning  of  all  scientific 
knowledge  of  electricity  and  electric  machinery. 

The  law  is  stated  as  follows: 

The  electric  current  along  a  conductor  equals  the  pressure 
divided  by  the  resistance. 

~  Pressure 

Current  =  ^ — :—  . 

Resistance 

In  electric  units: 

Volts 

Amperes  =  777- — . 
Ohms 

In  symbols: 

I==H' 


OHM'S  LAW  47 

Where  /  stands  for  Intensity  of  current  in  amperes. 
"      E     "       ll    Mectromotive  force  in  volts. 
"      R     '  '       tl    Resistance  in  ohms. 
The  student  will  save  himself  much  work  in  the  future 
if  he  will  take  the  trouble  to  learn  this  law  in  its  three  forms. 

pi 

1.  I  =~B:  Current  equals  voltage  divided  by  resistance. 

2.  E=IR:  Voltage  equals  current  times  resistance. 

El 

3.  R=-j-:  Resistance  equals  voltage  divided  by   current. 

Three  applications  of  this  law  are  given  to  illustrate  these 
three  ways  of  using  the  law.  A  more  complete  develop- 
ment of  its  use  will  follow. 

Example  1.     The  pressure  in  an  electric  circuit    is   50  volts, 
the  resistance  is  25  ohms.     How  many  amperes  flow  in  the  line? 

E  =  50  volts; 
72  =  25  ohms; 

E    50 
= 


Example  2.     How  many  volts  are  required  to  force  3  amperes 
through  20  ohms? 

7  =  3  amperes; 
72  =  20  ohms; 

=  60  volts. 


*v   Example  3.     Through  how  many  ohms  can  110  volts   force   5 
amperes? 

#=110  volts; 

7=5  amperes; 

R  =  T  =  -—-  =  22  ohms, 
I       o 


48  ELEMENTS  OF   ELECTRICITY 

Problem  4-3.     An  incandescent  lamp  uses  .5  ampere  on  a  110- 
volt  circuit.     What  is  the  resistance  of  lamp  when  burning? 

Problem  5-3.     If  72  =  11  ohms,  Fig. 
220  Volts  55,  how  many  amperes  are  flowing  in 

line? 


R 
AVWVWV 


Problem    6-3.     What     current     is 
produced    through    a    resistance    of    5 
Flo  55.  ohms   by    an    electromotive  force    of 

10  volts? 

Problem  7-3.  Through  what  resistance  will  an  electromotive 
force  of  15  volts  force  a  current  of  3  amperes? 

Problem  8-3.  What  voltage  will  produce  a  current  of  4  amperes 
through  a  resistance  of  4  ohms? 

Problem  9-3.  What  electromotive  force  will  produce  a  cur- 
rent of  .03  ampere  through  a  resistance  of  1000  ohms? 

Problem  10-3.  Through  what  resistance  will  121  volts  produce 
11  amperes? 

Problem  11-3.  What  current  is  produced  by  10  volts  acting 
across  .25  ohm? 

Problem  12-3.  A  dynamo  generates  500  volts;  the  resistance 
of  the  circuit  including  the  dynamo  is  20  ohms.  What  is  the 
current? 

Problem  1^3-3.  An  electric  bell  has  a  resistance  of  400  ohms 
and  will  not  ring  with  a  current  of  less  than  .03  ampere.  Neglect- 
ing battery  and  line  resistance,  what  is  the  smallest  electromotive 
force  that  will  ring  the  bell? 


50.  Applications  of  Ohm's  Law.  Ohm's  law  can  be 
applied  to  an  electric  circuit  as  a  whole,  or  it  can  be  applied 
to  any  part  of  it.  It  requires  a  great  amount  of  care  and 
practice  to  apply  this  simple  law  correctly  in  all  cases. 
Accordingly,  there  is  no  part  of  electrical  work  where  so 
many  mistakes  are  made  as  in  these  various  applications. 
Once  the  principle  is  firmly  grasped,  the  student  is  able 


OHM'S  LAW  49 

at  once"  to  attack  intelligently  a  wide  range  of  electrical 
problems. 

Many  of  the  difficulties  will  be  cleared  up  if  one  will  keep 
in  mind  the  two  following  statements  of  the  law  and  will 
use  them  intelligently: 

When  applying  the  law  to  an  entire  circuit,  state  the  law 
as  follows: 

(1)  The  current  in    the   Entire  circuit  equals   the  voltage 
across  the  Entire  circuit  divided  by  the  resistance  of  the  Entire 
circuit. 

Notice  that  the  word  ENTIRE  applies  to  CURRENT,  VOLTAGE, 
and  RESISTANCE  alike.  Not  to  one  only,  or  to  two  only, 
but  to  all  three  factors  of  the  equation. 

When  applying  the  law  to  but  a  part  of  the  circuit,  state 
the  law  as  follows : 

(2)  The  current  in  a  certain  Part  of  a  circuit  equals  the 
voltage  across  that  Same  Part,    divided  by  the  resistance  of 
that  Same  Part. 

Notice  here  again  that  the  values  for  the  three  facto-rs, 
CURRENT,  VOLTAGE,  and  RESISTANCE,  are  all  taken  from 
the  same  part  of  the  circuit.  By  far  the  greatest  number 
of  mistakes  in  applying  Ohm's  law  comes  from  dividing 
the  voltage  across  one  part  of  a  circuit  by  the  resistance  of 
another  part  and  expecting  to 
find  the  current  in  some  part  or 
other. 

A  few  examples  will  make  the 
correct  use  plain. 

Example.     The  generator  G,  Fig.  FIG.  56. 

56,  has  a  resistance  of  2  ohms  and 

generates  150  volts  pressure.  What  current  flows  through  a 
circuit  which  has  two  resistances  placed  in  series  with  the  genera- 
tor, if  one  resistance  is  8  ohms  and  the  other  10  ohms? 

Solution.  Since  the  pressure  generated  by  the  generator 
is  the  voltage  of  the  entire  system,  the  resistance  of  the  entire  sys- 
tem must  be  used  in  the  equation ; 


50  ELEMENTS  OF   ELECTRICITY 

voltage  (across  entire  system) 


Current  (m  entire  system)  . 

resistance  (of  entire  system) 

Voltage  (across  entire  system)  =  150  yolts; 
Resistance  (of  entire  system)    =2  +  8  +  10  =  20  ohms  ; 


Current  (in  entire  system)         =  ~~  =  ^'^  amPeres- 


That  is,   /  =  •=,    applied  to   the   entire  circuit.     Having   found 
R 

the  current  flowing  in  the  circuit,  it  is  now  possible  to  compute 
the  voltage  necessary  to  force  it  through  the  different  parts  by 
applying  Ohm's  law  to  that  part  alone. 

Suppose    it  is  desired  to  find  the  voltage    across    the  8-ohm 
resistance  alone. 

Voltage  (across  8  ohms)  =  current  (in  8  ohms)  X 

resistance  (of  8  ohms). 

Current  (through  8  ohms)  =  7.5  amperes. 
Resistance  (of  8  ohms)        =8  ohms. 

Voltage  (across  8  ohms)     =  7.5  X  8  =  60  volts. 
That  is, 

E  =  IR  applied  to  one  part  of  current  only. 

Across  the  10  ohms  resistance,  in  the  same  way, 

E=IR  =  7.5X10  =  75  volts; 
Across  the  interior  of  generator, 

#  =  7.5X2  =  15  volts. 

Total  voltage  =  60  +  75  +  15  =  150  volts,  which  checks  with  the 
value  given  as  the  voltage  generated. 

,61.  Brush    Potential  or    Terminal   Voltage.       We     may 

say  then,  that  in  the  above  example  75  volts  of  the  150  are 
used  to  force  the  current  through  the  10  ohms  resistance; 
60  volts  are  used  to  force  it  through  the  8  ohms  resistance, 
and  the  remaining  15  volts  are  used  to  force  it  through 
the  generator  itself. 


OHM'S  LAW  51 

The  pressure  used  to  force  the  current  through  the  two 
resistances  of  8  ohms  and  10  ohms,  would  be  the  75  volts, 
plus  the  60  volts  or  135  volts.  Since  this  135  volts  is  used 
entirely  in  forcing  the  current  from  one  brush  through 
the  outside  circuit  and  back  to  the  other  brush,  it  is  called 
the  BRUSH  POTENTIAL  or  TERMINAL  VOLTAGE  of  the  gen- 
erator for  the  given  current.  If  a  voltmeter  were  placed 
across  the  generator  it  would  indicate  the  TERMINAL  VOLT- 
AGE, since  that  is  all  the  voltage  available  for  forcing  a 
given  current  from  brush  to  brush,  through  the  external 
resistance.  This  BRUSH  POTENTIAL  of  a  generator  must 
always  be  distinguished  carefully  from  the  total  voltage 
generated  by  the  generator,  which  is  called  the  Electro- 
motive Force  of  the  generator.  This  is  generally  abbre- 
viated to  E.M.F. 

Thus  in  above  example, 

tbe  150  volts  equals  the  E.M.F. 
"    135     "      :•  "       "     Brush  Potential. 


Example.  Arc  lamp  A,  Fig.  57,  requires  5  amperes  "to  operate  it 
and  has  a  resistance  of  16  ohms.  There  is  a  resistance  B  of  4 
ohms  in  series  with  A. 

(a)  What  voltage  is  required  to  A 

operate  the  lamp? 

(6)  What  E.M.F.  must  a  genera- 
tor of  2  ohms  internal   resistance         VL^                            ^ 
develop  in  order  to  run  lamp  and  a         /                       B 
resistance  B  of  4  ohms?  ^~         — VWAAAAA/ ' 

(c)  What    is   voltage    across    B  FIG.  57. 
alone? 

(d)  What  is  the  Brush  Potential  of  the  Generator? 
Solution,     (a)  Voltage  across  lamp. 

/  =  5  amperes  (current  in  lamp) ; 

R  =  16  ohms  (resistance  of  lamp) ; 

#  =  7/2  =  5X16  =  80  volts  (pressure  across  lamp). 


52  ELEMENTS   OF   ELECTRICITY 

(6)  E.M.F.  generated  by  generator  G  (that  is,  of  course,  voltage 
across  entire  circuit). 

7  =  5  amperes  (current  in  entire  circuit)  ; 

22  ohms  (resistance  in  entire  circuit), 


E  =  IR  =  5X22  =  110    volts    (E.M.F.    generated    or    volts    across 
entire  circuit). 

(c)  Voltage  across  B. 

1  =  5  amperes  (current  in  7?); 
R  =  4  ohms  (resistance  of  B)  ; 
#  =  772  =  5X4  =  20  volts  (voltage  across  B). 

(d)  Brush  potential.     Brush  potential  is  voltage  across  all  that 
part  of  the  circuit  which  is  outside  of  the  generator. 

7  =  5  amperes  (current  through  circuit  outside  of  generator)  ; 
72  =  16  +  4  =  20  ohms  (resistance  outside  of  generator); 

£7  =  772  =  20X5  =  100  volts  (pressure  outside  of  generator  or  BRUSH 
POTENTIAL. 

Notice  that  in  each  case,  no  matter  what  the  problem,  the  method 
is  identical,  great  care  being  taken  in  using  the  equation  to  have 
the  current,  resistance,  and  voltage  include  exactly  the  same  part 
of  the  circuit  and  no  more. 

52.  Series  and  Parallel  Circuits.  There  are  two  ways 
of  connecting  two  or  more  pieces  of  electrical  apparatus 
together. 

(1)  SERIES.     When  the  pieces  are  connected  in  tandem 
they  are  said  to  be  in  SERIES.    Lamp  A  and  B,  of  Fig.  58, 
are  in  series. 

(2)  PARALLEL.     When  the  pieces  are  connected  so  that 
the  current  is  divided  between  them,  they  are  said  to  be 
in    PARALLEL  with  one  another.     MULTIPLE  or  SHUNT  are 
other  names  for  this  combination.     Lamps  C  and  D  of  Fig. 
59,  are  in  parallel  with  each  other. 


OHM'S    LAW  53 

The  two  combinations  may  exist  in  the  same  circuit, 
as  in  Fig.  60,  where  the  parallel  combination  C  and  D  is  in 
series  with  the  series  combination  A  and  B. 


FIG.  58. — Series  combination;  A  is        FIG.  59. — Parallel  combination;  C  is  in 
in  series  with  B.  parallel  with  D. 

Also  as  in  Fig.  61,  where  the  parallel  combination  C  and 
D  is  parallel  with  the  series  combination  A  and  B. 


FIG.  60. — Series  arrangement  of  parallel     FIG.  61. — Parallel  arrangement  of  series 
and  series  combinations.  and  parallel  combinations. 

53.  Series — Resistance,  Voltage,  and  Current  Relations. 

(1)  RESISTANCE.     The   resistance   of   a   series   combina- 
tion equals  the  sum  of  the  resistance  of  the  separate  parts. 
So  in  Fig.  58,  the  resistance  of  the  series  combination  of 
lamps  A  and  B,    equals  the  resistance  of  lamp  A  plus  the 
resistance  of  lamp  B. 

(2)  VOLTAGE.     The  voltage  also,  of  a  series  combination 
equals  the  sum  of  the  voltages  across  the  separate  resistances. 
Thus  the  voltage  across  the  lamps  A  and  B,  Fig.  58,  equals 
the  voltage  across  A  plus  the  voltage  across  B. 

As  has  been  stated  earlier  in  this  chapter,  the  voltage 
across  any  part  of  a  series  circuit  is  proportional  to  the 
resistance  of  that  part.  Thus  the  voltage  across  A  would 
be  3  times  the  voltage  across  B,  if  the  resistance  of  A  were  3 
times  the  resistance  of  B. 


54 


ELEMENTS  OF  ELECTRICITY 


(3)  CURRENT.     The   current   in   every   part   of   a   series 
circuit  is   the  same.     You  cannot  dam  up  an  electric  cur- 

rent. Therefore  the  current 
in  lamp  A,  Fig.  58,  must  be 
the  same  as  the  current  in 
lamp  B,  no  matter  what  their 
respective  resistances  and 
voltages  may  be.  The  volt- 

FIG.  62.  —  Series  combination.  ,  .    ,  , 

ages   and    resistances    always 
distribute  themselves  strictly  in  accordance  with  Ohm's  law, 


Example.     Series  combination  (Fig  62.)  : 

Resistance  A  equals  50  ohms. 
B       "     159    " 
C      "     .  7o    " 


Voltage  across  the  combination  equals  000  volts. 
Find: 

(1)  Combined  resistance  of  A,  J5,  C,  R. 

(2)  Current  through  each. 

(3)  Voltage  across  each. 

Solution. 

(1)  Resistance  of  combination: 

50  +  1  50  +  75  +  25  =  300  ohms. 

(2)  Current  through  each  : 

Apply  Ohm's  Law  to  the  combination  : 


Voltage  (across  combination) 
Current  (through  combination)-  Regista,.ee  (of  combination)  • 


600 


There  are,  therefore,  2  amperes  flowing  through  each  part  of 
the  combination. 


OHM'S   LAW  55 

(3)   Voltage  across  each. 

Apply  Ohm's  Law  to  each  part  of  the  combination  successively; 
to  lamp  A,  for  instance: 

E  =  IR. 
That  is, 

Voltage  (across  A)  =  current  (through  A)  X resistance  (of  A.) 

E    =2X50   =100  volts.  (Voltage  across  A). 

E'   =  2  X 150  =  300     ' '  (Voltage  across  B) . 

E"  =2X75   =150     "  (Voltage  across  C). 

#'"  =  2X25   =   50     "  (Voltage  across  #). 

E     (sum)      =  600     ' '  (Voltage  across  the  combination) . 

Problem  14-3.  What  brush  potential  must  a  generator  pro- 
duce to  supply  an  electroplating  current  of  20  amperes  through 
a  circuit  whose  total  resistance  is  .2  ohm? 

Problem  15-3.  A  dynamo  generates  an  electromotive  force  of 
1150  volts  and  delivers  a  current  of  20  amperes.  The  resistance 
of  the  circuit,  including  dynamo,  is  what? 

Problem  16-3.  (a)  If  the  magneto-generator  for  ringing  a  tele- 
phone bell  gives  an  electromotive  force  of  50  volts,  what  current 
will  be  transmitted  through  the  circuit?  The  resistance  of  gen- 
erator is  500  ohms,  of  line  and  bell  is  125  ohms.  (6)  What  will 
the  brush  potential  be? 

Problem  17-3.  What  E.M.F.  will  be  required  to  force  2  amperes 
through  a  series  circuit  containing  generator  of  £  an  ohm,  line  wires 
of  1^  ohms,  and  a  lamp  of  100  ohms  resistance?  (b)  What  will 
be  the  brush  potential  of  generator? 

Problem  18-3.  Lamp,  Fig.  63,  has  a  resistance  of  100  ohms, 
and  line  wires,  2  ohms  each,  (a)  What  is  current  in  line?  (6) 
What  is  voltage  across  the  lamp?  (c)  How  much  voltage  is  used 
up  in  sending  the  current  through  the  line  wires? 

Problem  19-3.  A  circuit  consists  of  8  ohms  resistance  in  gen" 
erator  and  3  ohms  in  line  wires,  and  has  30  lamps  in  series,  each 
of  8  ohms.  With  a  current  of  4  amperes,  what  will  be  the  drop 
in  the  generator;  (2)  in  the  line;  (3)  what  the  difference  of  potential 
across  each  lamp;  (4)  brush  potential  of  generator? 


56  ELEMENTS   OF    ELECTRICITY 

Problem  20-3.     (a)  If  each  lamp,  Fig.  64,  of  the  combination 
takes  .5  ampere,  how  many  amperes  must  the  generator  deliver? 

(6)  Resistance  of  ^4  =  100  ohms; 
"          "    5  =  200     " 
"    C  =  150     " 
Combined  resistance  =  ? 


2  Ohms 


2  Ohms 


)100 
Ohmsi 


FIG.  63.  FIG.  64. 

54.  Parallel — Voltage,  Current,  and  Resistance. 

(1)  Voltage.     The    voltage    across    each     branch    of    a 
parallel  circuit  is  the  same  as  the  voltage  across  the  com- 
bination.    In  Fig.  59,  the  voltage  across  C  is  the  same  as 
the  voltage  across  D,  which  is  the  same  as  the  voltage  across 
the  combination,  C  and  -D,  since  they  both  lie  between  the 
same  points,  X  and  Y. 

(2)  Current.     The    current    in     a    parallel    combination 
equals  the  sum  of  the  currents  in  the  separate  parts.     As 
the  current  flowing  from  X  to  Y  (Fig.  59),  equals  the  cur- 
rent in  C  plus  the  current  in  D. 

(3)  Resistance.     The   resistance  of    a   parallel    combina- 
tion equals  the  reciprocal  of  the  sum  of  the  conductances 
of  the  separate  parts. 

This  method  of  finding  the  combined  resistances  of 
pieces  in  parallel,  is  a  "  last  resort  "  method.  The  resistance 
can  generally  be  found  by  a  direct  application  of  Ohm's 
law,  but  sometimes  the  above  is  more  convenient. 

The  CONDUCTANCE  is  the  reciprocal  of  resistance;  that  is, 

Conductance  = 


Resistance' 

Thus   if  the  resistance  may    be   said  to  represent  the  dif- 
ficulty with  which  an  electric  current  is  forced  through  a  wire, 


OHM'S  LAW  57 

the  CONDUCTANCE  may  be  said  to  represent  the  ease  with 
which  the  current  can  be  forced  through  the  wire.  The 
name  "  Mho/'  the  inverse  of  "  Ohm,"  has  been  proposed 
as  the  unit  of  conductance.  A  wire  of  1  ohm  resistance  has  a 
conductance  of  1  mho.  A  wire  of  2  ohms  resistance  has  a 
conductance  of  J  mho,  etc. 


mho  = 


1 


ohm 


or     ohm  = 


1 


mho' 


Since  two  resistances  in  parallel  offer  less  RESISTANCE  to 
the  current  than  one  alone,  the  CONDUCTANCE  of  the  com- 
bination  is   greater  than   the 
conductance    of    one     alone. 
In  fact,  conductance  of  pieces 
in    parallel    equals    the    sum 
of  the  separate  conductances. 

Consider  Fig.  65. 

Resistance  of: 


FIG.  65. — Parallel  combination. 


L  =80  ohms; 
M=  8     " 
A=16     " 


Conductance  of: 


The  conductance  of  the  combination  therefore  isi 


or 


.0125  +.125  +.0025  =.20  mho, 


58  ELEMENTS  OF  ELECTRICITY 

If  the  conductance  of  the  combination  is  \  mho,  the  resistance 
of  the  combination  must  be  f  or  5  ohms,  since  the  resistance 
equals  the  reciprocal  of  the  conductance. 

Thus  the  rule  for  finding  the  resistance  of  a  parallel  com- 
bination is: 

Add  the  conductances  of  the  separate  branches,  which 
gives  the  conductance  of  the  combination.  Invert  the 
conductance  of  the  combination  to  get  the  resistance  of  the 
combination. 

Example.     In  Fig.  66,  (paral- 
lel combination.) 

Resistance  A  =  60  ohms ; 
."         £  =  40  ohms; 
C  =  80  ohms. 

Voltage    across    combination 
FIG.  66.  =  120  volts. 

Find: 

(1)  Voltage  across  each. 

(2)  Current  through  each. 

(3)  Current  through  combination. 

(4)  Resistance  of  combination. 

(1)  Voltage  across  each  is  120  volts,  because  in  parallel  combina- 
tions, the  voltage  across  each  branch  equals  the  voltage  across 
the  combination. 

(2)  Current  through  each. 

Apply  Ohm's  Law  to  each  branch  successively. 


K 

Voltage  (across  .4) 

Current  (through  A)  =— — : —       — ,  ,    .. — 
Resistance   (of  A) 

1  on 

Current  through  A  1= -—-  =  2  amperes ; 

oU 

120 
Current  through  B  ^=~7n~  =  ^  amPeres> 

120 

Current  through  C  /=—— -=1.5  amperes. 

oO 


OHM'S  LAW  59 

(3)  Current  through  combination  equals  the  sum  of  the  currents 
in  the  separate  branches : 

2  +  3  +  1.5  =  6.5  amperes. 

(4)  Resistance  of  combination : 
Apply  Ohm's  Law  to  the  combination: 

Voltage  (across  combination) 

Resistance  (ot  combination)  =-—  — — : — -. 

Current  (through  combination) 

120 

R  = =  18.46  ohms. 

o.5 

It  may  also  be  solved  by  the  "  inverse  of  the  sum  of  the  separate 
conductances,"    as  follows: 

1 — n+^^TjjJj;  mho,     (Conductance  of  combination.) 

oO      40      oO      *i4U 

Resistance  (of  combination)  =  - —  — . 

Conductance  (ot  combination) 


Therefore 


240 

——=  18.46  ohms.     (Checks  with  above  ans.) 

lo 


Problem  21-3.     If  each  lamp  of  the  combination  Fig.  67,  takes 
.5  ampere,  how  many  amperes  must 
the  generator  deliver? 

Problem  22-3.  A  divided  circuit 
has  two  branches  of  1  ohm  and  i 
ohm  respectively.  What  is  the  joint 
conductance  of  the  two  branches?  FIG.  67. 

What  is  the  joint  resistance? 

Problem  23-3.  A  circuit  has  three  branches  of  12,  4,  and  6 
ohms  respectively.  If  4  amperes  flow  in  the  circuit  containing 
6  ohms,  what  current  will  flow  in  each  of  the  others? 

Problem  24-3.  What  pressure  will  be  required  to  force  10 
amperes  through  a  parallel  combination  consisting  of  4  ohms, 
5  ohms,  and  8  ohms? 

55.  Simple  Parallel  Lighting  Systems.  The  modern 
incandescent  lamps  are  usually  installed  in  parallel.  The 


60  ELEMENTS  OF  ELECTRICITY 

resistance  of  all  the  lamps  of  any  make  is  not  the  same. 
Nor  is  the  voltage  across  all  the  lamps  when  installed  the 
same..  Still  for  convenience  in  calculating  the  "  line  drop," 
efficiency  of  transmission,  etc.,  each  lamp  is  assumed  to 
take  the  same  current.  The  error  introduced  by  this  assump- 
tion is  usually  too  small  to 
be  taken  into  account. 

A  general  problem  may 
be  stated  as  follows : 

A  lighting  circuit  is  ar- 
ranged with  two  groups  of 
lamps  as  in  Fig.  68.  One 

group  has  two  lamps,  the  other,  three.     Each  lamp  takes  2 
amperes.     The  generator  has  a  terminal  voltage  of  120  volts. 


Resistance  of  line  wire  AB  and  DE=.6  ohm  each. 
"         "     "       "    BCand  EF=.3  ohm  each. 


Find: 


(a)  Volts  lost  in  line  wires. 

(6)  Voltage  across  each  group  of  lamps. 


First.  Find  the  current  distribution  throughout  circuit, 
assuming  that  each  lamp  takes  2  amperes. 

Current  through  BC  =   4  amperes 
«  .«       EF=  4       " 

"       AB=W       " 
"  "       DE=1Q       " 

Second.     By  Ohm's  law,   solve  for  the  "  drop  "  in  line 
wire. 

Drop  (in  AB)  =current  (through  A B)  X resistance  (of  AB). 

=  .6X10=6  volts. 
Drop  (in  DE)  =  current  (through  DE)  X  resistance  (of  DE), 

=  .6X10=6  volts. 


OHM'S  LAW  61 

Total  volts  used  in  line  between  generator  and  first  group 
of  lamps  then  equals, 

6+6=12  volts. 

Generator  voltage  -  voltage  lost  in  lead  wires  =  voltage 
at  lamps  (BE). 

120-  12  =108  volts  across  BE. 

Volts  (lost  in  BC)  =  current  (through BC)x resistance  (of  BC) ; 

=4X.3=1.2; 
Volts  (lost  in  EF)  =4  X  .3  =  1 .2 ; 

Total  lost  between  two  groups  of  lamps  =1.2+1.2  =2.4  volts. 
Voltage   at   first   group  —  volts   lost   in   connecting  wires  = 

voltage  at  second  group. 

108-2.4  =  105.6  volts  across  CF. 
Thus: 

The  voltage  across  first  group  =108     volts. 
"    second  "     =105.6  volts. 

The  above  simple  method  by  which  this  general  problem 
was  solved,  can  be  used  in  solving  any  problem  in  Direct 
Current  Light  and  Power  distribution. 

56.  Standard  Units.  So  far,  only  the  fundamental 
ideas  of  volts,  amperes  and  ohms  have  been  set  forth.  It 
is  essential,  however,  that  the  student  know  what  the  stand- 
ard value  of  each  is,  and  how  it  is  obtained. 

It  is  necessary  to  define  independently  but  two  of  the 
three  quantities.  The  third  may  be  determined  by  the 
application  of  Ohm's  law.  It  is  customary  at  the  present 
time  to  determine  independently  the  Ohm  and  the  Ampere, 
and  then  define  the  Volt  as  that  pressure  which  will  cause 
one  ampere  to  flow  through  one  ohm.  Probably,  a  better 
method  is  to  define  independently  the  Ohm  and  the  Volt, 


62  ELEMENTS  OF  ELECTRICITY 

and  set  the  value  of  the  Ampere  as  the  current  which  one 
volt  pressure  will  force  through  one  ohm  resistance.  The 
reason  why  the  second  method  is  to  be  preferred  is  that 
the  volt  can  be  determined  independently  much  more 
accurately  than  can  the  ampere. 

THE  STANDARD  OHM  is  the  resistance  of  a  column  of  pure 
mercury  106.3  cms.  long,  of  uniform -cross-section,  and  weigh- 
ing 14.4521  gms.  at  0°  C. 

THE  STANDARD  VOLT  is  defined  as  of  the  voltage 

of  a  Standard  Weston  cell.  This  latter  definition  has  been 
adopted  on  account  of  the  difficulty  of  maintaining  and  re- 
producing the  standard  conditions  in  solutions  of  silver 
nitrate. 

THE  STANDARD  AMPERE  is  the  rate  of  flow  of  a  steady 
current  which  one  STANDARD  VOLT  pressure  forces  through 
one  STANDARD  OHM  resistance.  The  theory  of  these  standard 
measurements  is  taken  up  more  fully  in  Chapters  V  and  XII. 


OHM'S  LAW  63 


SUMMARY  OF  CHAPTER  III 

Electricity  is  said  to  flow;  thus  we  speak  of  an  electric 
current. 

UNITS 

AMPERE.  (Rate  of  flow).  One  coulomb  per  second, 
or  current  which  will  deposit  in  a  given  time  a  standard 
amount  of  silver  from  standard  silver  nitrate  solution. 

OHM.  (Resistance  to  flow.)  Resistance  offered  by  standard 
column  of  mercury. 

VOLT.  (Pressure  causing  current  to  flow).  Pressure  to 
cause  one  ampere  to  flow  through  one  ohm  resistance. 

POTENTIAL  DIFFERENCE.  The  fundamental  requisite 
for  a  flow  of  current  in  an  electric  circuit;  can  be  called  the 
electric  difference  in  level  between  two  points.  The  cur- 
rent flows  from  the  higher  to  the  lower  level;  potential 
difference  is  measured  in  volts. 

OHM'S  LAW. 

volts  E 

Amperes  =  — — ,  I=^r- 

ohms  R 

Volts       =  amperes  X  ohms,  E=IR. 

volts  E 

Ohms      = ,  R=T- 

amperes 

Ohm's  law  should  be  understood  to  mean  always: 

The  current  in  a  certain  part  of  the  circuit  equals  the 
voltage  across  that  same  part  divided  by  the  resistance  of  that 
same  part. 

SERIES  CIRCUITS.  Combined  resistance  of  parts  in  series 
equals  sum  of  the  separate  resistances. 

Combined  Voltage  across  parts  in  series  equals  sum  of 
voltages  across  the  separate  resistances. 

Combined  Current.    Current  is  same  in  each  part. 

PARALLEL  CIRCUITS.  Combined  resistance  of  branches 
equals  the  reciprocal  of  the  sum  of  the  separate  conductances. 

Combined  current  equals  the  sum  of  the  currents  in  the 
separate  branches. 

Combined  Voltage.  Voltage  across  each  branch  is  the  same 
as  the  voltage  across  the  combination. 

Conductance  is  the  reciprocal  of  resistance. 


64  ELEMENTS  OF  ELECTRICITY 


PROBLEMS,    CHAPTER   III 

25-3.  If  a  car  heater  is  supplied  with  a  pressure  of  550  volts 
from  the  trolley,  how  great  must  its  resistance  be  that  the  current 
may  not  exceed  5  amperes  ? 

26-3.  A  16-c.p.  lamp  requires  .5  ampere,  and  its  resistance 
at  full  candle-power  is  200  ohms.  What  voltage  must  be  impressed 
on  its  terminals? 

27-3.  Find  drop  across  wire  of  78.8  ohms  resistance,  if  200 
amperes  are  forced  through  it. 

28-3.  Through  M,  Fig.  69,  5  amperes  flow.  Through  M' 
15  amperes.  Total  amperes  in  line  equal  what? 


FIG.  69. 


29-3.  If  each  lamp,  Fig.  70,  takes  .4  ampere,  how  much  cur- 
rent flows  in  the  following  sections  of  the  line:  AB,  BC,  DE,  and 
EF? 

30-3.  What  pressure  is  needed  for  an  incandescent  lamp  of 
50  ohms  resistance,  through  which  flows  a  current  of  1.04  amperes? 

31-3.  A  parallel  circuit  has  resistances  in  the  several  branches 
of  1,  2,  4,  5,  and  10  ohms  respectively.  What  is  the  conductance 
of  the  combination?  What  is  the  resistance? 

32-3.  Through  each  lamp,  Fig.  71,  4  amperes  flow. 

Find  current  from  A  to  B; 
11     "  BtoF; 

"     "  BtoC', 

"     "  CtoE. 

33-3.  A  circuit  has  two  branches,  one  of  2  ohms  with  a 
current  of  6  amperes,  the  other  of  15  ohms.  What  current  flows 
in  the  second  branch  and  what  difference  of  potential  is  main- 
tained between  the  terminals  of  the  circuit? 


OHM'S  LAW 


65 


34-3.  Lamp  L,  Fig.  72,  requires  2  amperes.  Brush  potential 
of  generator  is  220  volts  : 

(a)  How  many  volts  are  used  to  send  current  through  the  line 
wires? 

(6)  What  voltage  is  there  across  the  lamp? 

(c)  What  is  the  resistance  of  the  lamp? 


4  Ohms 


4  Ohms 


' 


- 


FIG.  71. 


FIG.  72. 


35-3.  A  5000-ohm  galvanometer  has  its  terminals  connected 
to  two  points  in  a  circuit  between  which  there  is  a  difference 
of  potential  of  .01  volt.  What  current  flows  through  the  gal- 
vanometer? 

36-3.  Three  resistances  of  20,  15,  and  8  ohms  are  in  series, 
with  200  volts  applied  across  the  outside.  What  will  be  the  dif- 
ference of  potential  across  each  resistance?  Current  through  each? 

37-3.  There  are  seven  arc  lamps  in  series,  Fig.  73,  each  requiring 
5  amperes.  If  each  has  a  resistance  of  16  ohms,  how  many  volts 
must  dynamo  supply  to  system? 


.25  Ohms 


.25  Ohms 


FIG.  74. 


38-3.  Motor  requires  20  amperes  at  110  volts,  in  Fig.  74. 
Line  wires  have  .25  ohm  each.  What  pressure  must  be  supplied 
by  generator? 

39-3.  A  wire  40  ft.  long  has  a  drop  of  2  volts  across  it.  What 
will  be  the  drop  across  20  ft.? 

40-3.  It  is  desired  to  find  the  resistance  of  the  motor  in  Fig.  75. 
The  motor  (Af)  is  put  in  a  circuit  in  series  with  a  resistance  (R) 
of  .020  ohm.  The  drop  across  #  =  .436  volt;  the  drop  across 
M  =  .348  volt.  What  is  the  resistance  of  the  motor? 


ELEMENTS  OF  ELECTRICITY 


41-3.  Each  of  the  8  arc  lamps  in  Fig.  76  requires  6  amperes. 
The  ordinary  resistance  of  each  lamp  is  13  ohms.  They  are  put 
8  in  series  on  an  800-volt  line.  How  much  extra  resistance  must 
be  added  to  each  lamp  to  cut  the  current  down  to  the  required 
6  amperes? 


42-3.  The  average  resistance  of  each   lamp  in  Fig.  77  is  220 
ohms.     Each  lamp  takes  .3  ampere, 
(a)  What  is  voltage  across  lamps? 
(6)  Voltage  lost  in  line? 
(c)  What  is  brush  potential  of  the  generator? 


2  Ohms 


43-3.  Each  lamp  in  Fig.  78,  takes  .5  ampere. 

(a)  What  is  voltage  across  CD  and  EFt 

(6)  What  is  average  resistance  of  each  lamp  across  CD? 

(c)  Across  EFt 


Volts  lost  in  line? 


1000  ft.— >j 


6X  i 


FIG.  79. 


FIG.  80. 


44-3.  Generator  D,  Fig.  79,  has  an  E.M.F.  of  115  volts  on  open 
circuit;  lamps  have  a  resistance  of  220  ohms  each.  Generator 
is  able  to  supply  .5  ampere  to  each  lamp. 

(a)  What  is  brush  potential  when  supplying  this  current? 

(6)  What  is  the  internal  resistance  of  dynamo?  (Neglect 
feed- wire  resistance.) 


OHM'S  LAW  67 

45-3.  A  storage  battery  has  an  E.M.F.  of  2.20  volts  on  open 
circuit;  the  internal  resistance  is  .002  ohm.  What  is  the  terminal 
voltage  when  it  is  delivering  10  amperes? 

46-3.  Each  lamp  in  Fig.  80  is  to  take  1  ampere.  There  are 
available  for  feed  wires  two  coils,  one  having  a  resistance  of  2 
ohms,  the  other  3  ohms;  each  coil  contains  2000  ft. 

(a)  Should  the  3-ohm  lot  be  used  between  the  dynamo  and 
group  A,  or  between  group  A  and  group  B? 

(6)  What  is  the  greatest  voltage  that  can  be  obtained  across 
"14  and  across  B,  using  these  two  coils  as  lead  wires? 

(c)  What  difference  in  the  voltage  across  A  would  a  poor  use 
of  the  wire  make? 

47-3.  Two  resistances,  150  and  100  ohms  respectively,  are 
connected  in  parallel  between  two  points  A  and  B,  across  which 
the  pressure  is  110  volts. 

(a)  What  current  will  flow  in  each  circuit? 

(fr)  If  the  100-ohm  circuit  be  reduced  to  2  ohms,  what  current 
will  flow  in  each? 

(c)  What  is  the  resistance  of  the  parallel  combination  in  each  case? 

Resistance  of  A  =  100  ohms. 


B  =  120  ohms. 
(7  =  160  ohms. 
Find: 

(a)  Current  through  each  resistance. 
(6)   Resistance  of  parallel  combination  (A 
and  B). 

(c)  Combined  resistance  of  system. 

(d)  Voltage  across  each  resistance.  FiG.*8i. 


49-3.  In  Fig.  82,  Voltage  from  A  to  5  =  40  volts. 
Current  through  resistance  x  is  2.5  amperes. 
Resistance  y  =  5  ohms. 
2  =  4  ohms. 
Find: 

Current  through  y.     Resistance  of  x. 

z.     Voltage  from  B  to  (7. 

50-3.  A  lighting  circuit  contains  65  arc  lamps  in  series,  each 
of  8  ohms  resistance.  If  the  resistance  of  the  generator  is  4  ohms 
and  of  the  line  is  8  ohms,  what  electromotive  force  must  be  main- 


68 


ELEMENTS   OF   ELECTRICITY 


tained  to  supply  a  current  of  3  amperes?     What  is  the  brush 
potential  of  the  generator? 

51-3.  In  Fig.  83,  M\  has  a  resistance  of  5.5  ohms,  and  requires 
a  current  of  20  amperes;    M%  requires  10  amperes, 
(a)  What  voltage  across  MI  and  across  If  2? 
(6)  What  is  the  combined  resistance  of  AC  and  BD? 

.08  Ohm         A 


.12  Ohm 


Fro.  83. 


62-3.  Each  lamp,  Fig.  84,  has    220  ohms  resistance,  and  is 
run  on  110  volts. 
Find: 

(a)  Total  current  through  lamps. 
(6)  Volts  lost  in  line. 

(c)  Brush  potential  of  generator. 

(d)  E.M.F.  of  generator < 

(e)  Volts  lost  in  generator. 


2  Ohms 


-.24 


.24 


FIG.  84. 


Fro.  85. 


63-3.  In  Fig.  85,  voltage  across  group  .4  =  116  volts.     Resist- 
ance of  each  lamp  in  group  A  =  100  ohms. 
Find: 

(a)  Voltage  across  group  B. 

(b)  Average  resistance  of  each  lamp  in  group  B. 

(c)  Line  drop  between  generator  and  group  A. 

(d)  Line  loss  in  volts  between  A  and  B. 

(e)  Current  through  each  lamp  in  A  and  each  lamp  in  B. 

64-3.  What  would  the  values  in  Problem  53  become  if  the 
resistance  from  G  to  A  were  .3  ohm  for  each  wire,  and  from  A  to 
B  .24  ohm  for  each  line  wire? 


OHM'S  LAW 


69 


65-3.  A  circuit  consists  of  the  following  4  parts  joined  in  series. 
First  part:  2  coils  joined  to  each  other  in  parallel,  one  having  4 
ohms  resistance,  the  other  7  ohms.  Second  part :  a  line  of  wire  of 
16  ohms  resistance.  Third  part:  12  lamps  in  parallel,  each  hav- 
ing 220  ohms  resistance.  Fourth  part:  a  line  wire  of  2  ohms 
resistance. 

What  voltage  is  required  to  send  5  amperes  through  this  circuit? 

56-3.  In  Fig.  86, 


FIG.  86. 


Car  No.  1  is  1  mile  from  station  and  is  taking  40  amperes. 
"       2 "  3  miles         "  "  20 

3 «  4  «  a  25        " 

Trolley  has  a  resistance  of  .42  ohm  per  mile.     Track  resistance 
.03  ohm  per  mile. 
Find: 

(a)  Voltage  across  each  car. 
(6)  Total  voltage  loss  in  line. 

57-3.  What  would  the  values  in  Problem  56  become  if  a  feeder 
of  .26  ohm  per  mile  were  run  along  the  trolley,  being  tied  to  it  every 
half  mile? 


CHAPTER  IV 
POWER  MEASUREMENT 

Use  of  Ammeter,  Voltmeter,  and  Wattmeter — Electric  Power;  Watt; 
Kilowatt — Variations  of  Power  Equation — Electric  Energy; 
Kilowatt-hour;  Watt-second  or  Joule — Electric  Energy  Converted 
to  Heat  Energy— Heat  Equivalent  of  Electricity— Efficiency:  of 
Electric  Machines;  of  Electric  Transmission;  of  Electric  Lamps. 

67.  Ammeter.  When  we  wish  to  know  how  many  am- 
peres are  flowing  through  a  given  part  of  a  circuit,  we  insert 
an  ammeter  in  that  part  of  the  circuit.  Since  an  ammeter 
reads  only  the  current  going  through  itself,  we  must  actually 
cause  the  current  we  wish  to  measure  to  flow  through  the 
instrument.  This  means,  we  must  break  the  circuit  and 
insert  the  ammeter  in  such  a  way  that  the  current 
flows  through  it.  In  other  words,  an  ammeter  is  always 
placed  in  series  in  a  circuit. 

In  Fig.  87,  ammeter  A  measures  the  current  through 
R  because  it  is  placed  in  series  with  R.  Since  the  circuit 


FIG.  87. — Ammeter  A  measures  FIG.  88. — Ammeter  A  measures 

current  in  R  and  A".  current  in  M  only. 

is  a  simple  series  circuit,  the  ammeter  also  measures  the 
current  in  arc  L  and  generator  G.  Notice  that  the  circuit 
is  broken  to  insert  ammeter. 

In  Fig.  88,  the  ammeter  A  measures  the  current  flowing 
through  the  motor  M,  because  it  is  in  series  with  the  motor. 

70 


POWER   MEASUREMENT  71 

It  does  not,  however,  measure  the  current  in  the  lamp  L, 
because  it  is  not  in  series  with  the  lamp.  Neither  does  it 
measure  the  current  through  the  generator  G.  The  cur- 
rent through  the  generator  must  equal  the  current  through 
motor  M  plus  the  current  through  lamp  L,  as  explained 
in  Chapter  III. 

Since  an  ammeter  must  always  be  of  very  low  resistance 
in  order  not  to  cut  down  the  current,  it  is  necessary  always 
to  use  a  short-circuiting  switch  with  the  instrument.  This 
protects  it  from  injury  in  case  there  is  too  much  current 
flowing  through  the  line  for  the  instrument  to  record. 

In  Fig.  89,  with  the  short-circuiting  switch  in  position 
S,  nearly  all  the  current  flows  through  the  switch-blade.  If 
s' 


FIG.  89. — Ammeter  with  short  circuiting  switch.  FIG.  90. 

there  is  too  much  current  in  the  line  for  the  instrument  to 
carry,  there  will  be  a  noticeable  reading  on  the  instrument 
with  switch  S  closed.  If,  however,  the  ammeter  registers 
nothing,  it  will  probably  be  safe  to  throw  the  switch  to  the 
position  S',  which  causes  all  the  current  to  flow  through 
the  ammeter,  where  it  will  be  indicated. 

Care  must  be  taken  to  get  the  +  side  of  the  instrument 
on  the  +  side  of  the  line. 

58.  Voltmeter.  Pressure  between  two  points  is  measured, 
by  placing  a  voltmeter  across  these  points.  A  voltmeter 
is  not  inserted  in  a  circuit,  but  merely  placed  in  shunt  with 
the  part  of  the  circuit  across  which  it  is  desired  to  find  the 
voltage. 

In  Fig.  90,  if  it  is  desired  to  find  the  voltage  across  R, 
the  voltmeter  is  connected  as  shown,  to  the  two  ends  of  R, 
without  disturbing  the  former  connections.  Be  sure  to 
get  -f  side  of  voltmeter  on  the  +  side  of  the  line.  A 


72  ELEMENTS  OF  ELECTRICITY 

voltmeter  is  of  very  high  resistance,  and  therefore  takes 
no  appreciable  amount  of  current  from  the  line. 

Just  as  an  ammeter  indicates  the  current  flowing  through 
itself,  so  a  voltmeter  reads  the  voltage  across  itself.  Thus 
in  Fig.  90,  when  we  wish  to  measure  the  voltage  across  the 
resistance  R,  we  place  the  voltmeter  in  parallel  with  R, 
so  that  the  voltage  across  it  will  be  the  same  as  the  voltage 
across  R.  Therefore  when  it  indicates  the  voltage  across 
itself,  it  also  indicates  the  voltage,  across  R. 

59.  Rule  for  Use  of  Ammeter  and  Voltmeter.  Place 
AMMETER  in  series,  always  using  a  short-circuiting  switch 
to  prevent  injury  to  the  instrument. 

Place  VOLTMETER  in  shunt.  Put  the  -f  side  of  each 
instrument  on  the  -f-  side  of  the  line. 

Fig.  91  shows  the  correct  use  of  an  ammeter  and  a  volt- 
meter to  measure  the  current  and  the 
voltage  supplied  to  the  motor  M. 

The  short-circuiting  switch  S  must  be 
opened  before  the  ammeter  is  read. 

All  the  current  that  enters  the  motor 
must  then  flow  through  the  ammeter  and 
be  indicated.  The  ammeter  is  of  very 
low  resistance  (about  .001  or  .002  ohm) 
and  does  not  cut  down  the  flow  of  the 
current. 

The  voltmeter  is  of  very  high  resistance 
(about  15,000  ohms),  and  does  not  allow  any 
appreciable  current  to  flow  around  the 
motor,  yet  enough  goes  through  it  to  cause 
it  to  indicate  the  voltage  across  the  terminal  A B  of  the 
motor. 

Suppose  the  voltage  across  the  motor  to  be  110  volts, 
what  would  happen  if  an  ammeter  of  .002  ohm  resistance 
were  by  mistake  placed  across  AB1  (Remember — 
Ohm's  law  is  always  in  operation.)  For  a  detailed  dis- 
cussion of  ammeters  and  voltmeters  see  Chapter  XIV. 


POWER  MEASUREMENT  73 


WORK  AND  POWER  MEASUREMENTS 

60.  Power,  Watt,  Kilowatt.  The  flow  of  an  electric 
current  has  been  likened  to  the  flow  of  water  through  a 
pipe.  A  current  of  water  is  measured  by  the  number  of 
gallons  or  pounds  flowing  per  minute  ;  a  current  of  electricity, 
by  the  number  of  amperes,  or  coulombs  per  second.  The 
power  required  to  keep  a  current  of  water  flowing  is  the 
product  of  the  current  in  pounds  per  minute  by  head,  or 
pressure,  in  feet.  This  gives  the  power  in  foot-pounds  per 
minute.  To  reduce  to  horse-power,  it  is  necessary  merely 

r   -j    u    oo  nnn   •       feet  X  (Ibs.  per  min.) 
to  divide  by  33,000,  i.e.,  -  -  =  H.P. 


In  exactly  the  same  way,  the  power  required  to  keep  a 
current  of  electricity  flowing,  is  the  product  of  the  current 
in  amperes  by  the  pressure  in  volts.  This  gives  the  power 
in  watts.  To  reduce  to  Kilowatts  it  is  necessary  merely 
to  divide  by  1000,  i.e. 

volts  X  amperes  _  T  ,  w 
1000 

1  Kilowatt        =1.34  Horse-power  or  1£  H.P. 
1  Horse-power  =746  watts  or  f  Kilowatt. 
This  method  of  computing  electric  power  may  be  written 
in  the  form  of  an  equation  as 

P=IE. 

where  P=  power  in  Watts' 

I  =  current  in  Amperes; 
E  =  pressure  in  Volts. 

The  two  following  examples  show  the  similarity  in  the 
methods  of  computing  water  power  and  electric  power: 


74  ELEMENTS  OF   ELECTRICITY 

Example.  What  power  is  required  to  drive  a  pump  which 
must  raise  water  100  ft.  and  supply  1500  gals,  per  hour?  (1  gal. 
of  water  weighs  8.31bs.) 

Power   =  pressure  X  current  ; 

=  feet  X  pounds  per  min.; 

=  100  ft.X1500*8'3  Ibs.  per  min; 
oO 

-20,800  ft.  Ib.  per  min.; 

20  800 

20,800  ft.-lb.  per  min.  =  -^   ~  H.P.  =  .63  H.P. 
00,000 

About  a  1  H.P.  motor  would  then  be  required  to  drive  the  pump, 
when  delivering  the  current. 

Example.  What  power  is  required  to  drive  a  generator  which 
must  deliver  4  amperes  at  120  volts? 

Power  =  pressure  X  current. 
=  volts  X  amperes. 
=  120X4  =  480  watts  or  .480  K.W. 

480  watts  =  |?5  H.P.  =  .64  H.P. 


Thus  a  motor  of  somewhat  less  than  a  1  H.P.  would  be  re- 
quired to  drive  the  generator  when  delivering  this  current. 

Note  that  these  methods  of  computing  the  power  are  identical. 

61.  To  Compute  the  Power  in  an  Electric  Circuit.      If 

then  we  wish  to  know  the  power  that  is  being  consumed  in 
a  certain  part  of  an  electric  circuit,  we  merely  have  to  insert 
an  ammeter  to  measure  the  current  in  that  part  of  the  cir- 
cuit, and  apply  a  voltmeter  to  measure  the  voltage  across 
that  part  of  the  circuit,  and  multiply  the  Ammeter 
readings  by  the  Voltmeter  reading.  This  gives  us  our 
power  directly  in  Watts.  This  is  usually  stated  as 
an  equation  as  follows: 

Watts  =  Volts  X  Amperes. 

The  same  precautions  must  be  observed  in  the  use  of 
this  equation  as  in  the  use  of  Ohm's  law.     That  is,    the 


POWER  MEASUREMENT  75 

Voltage  and  the  Current  must  be  measured  for  the  same 
part  of  the  circuit  at  the  same  time,  and  their  product  is 
the  power  consumed  in  that  part  of  the  circuit  alone.  The 
following  example  illustrates  the  use  of  the  equation. 

Example.     A  generator  G,  Fig.  92,  is  furnishing  a  current  of 
4  amperes  to  the  line  at  a  pres- 
sure  of  120  volts.     There  is  in  the 
circuit  a  resistance  R,  which  re-      > 
quires  5  volts  to  force  the  current 
through   it,    and     a     motor    M, 
which  requires  115  volts.     How 

much  power  does  the  resistance  R  consume,  and  how  much 
does  the  motor  M  consume? 

The  resistance  R  consumes  : 


or 

P(watts   in   R)=  /(current   through   R)  X  ^(voltage   across   R). 
P  =  4(amperes.)  X  5  (volts)  =-20  watts. 

Motor  M  consumes  : 

P  =  IE, 

or 

P(watts  in  M}=  /(current  through  M)X  /?  (voltage  across  M). 
P  =  4(amperes)X115  (volts)  =460  watts. 

The    total    power    consumed    by    R    and    M  =  460  +  20  =  480 
watts.     • 

This  checks  with  the  total  power  delivered  to  the  circuit  as 
follows: 

P  =  IE(for  circuit), 
or 

P(watts    in    circuit)  =/  (current    in    circuit)  XE  (voltage    across 
circuit)  . 

P  =  4  (amperes)  X  120  (volts)  =480  watts. 

62.  Variations  of  Power  Equation.     Just  as  Ohm's  law 
has  the  three  forms,   (1)   /=-|,  (2)  E=IR,   (3)  #=j,  so 


76  ELEMENTS  OF  ELECTRICITY 

this  power  equation  P=IE  may  have  three  forms,  found 
as  follows: 

(1)  P=IE; 
but, 

E=IR.  Ohm's  law. 

Therefore, 

P=I(IR)  or  PR,  substituting  in  (1). 

(2)  P=I2R; 
again, 

ci 

/  =-=5*  Ohm's  law. 

H 

Therefore, 

— J#=— ,  substituting  in  (1). 

E2 

(3)  P-f. 

It  is  well  to  learn  this  equation  in  its  three  forms: 


This  will  save  a  considerable  amount  of  mathematical 
work. 

When  the  volts  and  amperes  are  given,  multiply  direct 
to  get  the  watts  (P=IE);  when  the  amperes  and  ohms 
are  known,  multiply  ohms  by  square  of  amperes  (P=I2R}\ 
when  volts  and  ohms  are  known,  divide  square  of  volts 

/       E2\ 
by  ohms  (P  =-p-J. 


The  result  is  the  same  as  though  we  used  Ohm's  law  first 
to  find  the  amperes  and  volts  and  then  multiplied. 


POWER  MEASUREMENT  77 

Example.     Generator  G,  Fig.  93,   furnishes  .5  ampere  to  the 
line  at  115  volts  pressure.     What 

power   is   consumed    by    (a)  the  f~  ^L  \~zfi1 

110-  volt  lamp  L,  and  by  (6)  the  ||  rcM  i_6  §\ 

10  ohms  resistance  R?  S[  ^~/  R 

(a)  For  the  lamp  L  :  ~*—^-  -  WAAA^  -  U=* 
VllO  volts; 
/  =  .  5  ampere;  FlG"  93' 


(1)  P  =  #/  =  .5X110  =  55  watts. 
Or  find  the  resistance  of  the  lamp. 

R=  |L  1^  =  220  ohms, 
1       .5 

Then 

/  =  .5  ampere  ; 
R  =  220  ohms; 

(2)  P  =  PR  =  .5  X  .5  X  220  =  55  watts. 

Or 

#  =  110  volts; 
jR  =  220  ohms. 

- 


The  result  of  any  of  the  three  methods  is  the  same,  55  watts. 
Therefore  it  is  always  best  to  use  the  easiest  method  for  the  data 
given.  Since  the  voltage  and  the  current  in  this  case  were  given, 
it  is  best  to  use  P  =  EI. 

(b)  For  the  resistance  R. 

R  =  W  ohms; 
/  =  .5  ampere  ; 
P  =  /2fl  =  .5X.5XlO=2.5  watts. 

There  is  no  need  of  finding  the  voltage  across  R,  and  multiply- 
ing that  by  the  current,  though  the  result  would  be  the  same. 
Thus 

115  volts  —  110  volts  =5  volts  across  R. 

E  =  5  volts; 
/  =  .5  ampere  ; 

2,5  watts, 


78  ELEMENTS  OF  ELECTRICITY 

Problem  1-4.  How  many  horse-power  are  required  to  hoist 
a  load  of  2000  Ibs.  40  ft.  per  minute? 

Problem  2-4.  How  many  kilowatts  are  required  to  supply 
the  power  in  Problem  1-4? 

Problem  3-4.  What  power  is  being  used  by  an  incandescent 
lamp  which  takes  .5  amperes  at  110  volts? 

Problem  4-4.  What  power  does  a  car  heater  use  if  its  resistance 
is  100  ohms  and  the  pressure  across  it  is  550  volts? 

Problem  5-4.  What  power  does  an  electric  flatiron  use  on  a 
11 5- volt  circuit  if  the  resistance  is  50  ohms? 

Problem  6-4.  How  many  watts  are  consumed  in  lead  wires 
in  Fig.  72? 

Problem  7-4.  A  110- volt  arc  lamp  requires  6  amperes  to  operate 
properly.  What  power  does  such  a  lamp  take? 

Problem  8-4.  How  many  watts  are  used  by  motor  Mt  in 
Fig  83?  How  many  watts,  by  the  line  wires  between  D  and 
M2? 

Problem  9^4.  What  power  is  consumed  by  each  lamp  in  Group 
A,  of  Fig.  85? 

Problem  10-4.  The  power  being  delivered  by  the  generator 
in  Fig.  86  is  equivalent  to  how  many  horse-power? 

63.  Use  of  Wattmeter.  Instead  of  using  two  separate 
instruments,  an  ammeter  and  a  voltmeter,  to  measure  the 
power  consumed  in  a  certain  part  of  a  circuit,  we  may  use 
a  single  instrument  called  a  WATTMETER.  This  instrument 
is  a  combination  of  an  ammeter  and  a  voltmeter,  and  thus 
has  an  ammeter  part  of  low  resistance  to  measure  the  cur- 
rent, and  a  voltmeter  part  of  high  resistance  to  take  the  volt- 
age. The  indicator  shows  the  product  of  the  volts  times 
the  amperes,  that  is,  the  watts. 

A  wattmeter,  then,  usually  has  four  terminals,  two  for 
the  ammeter  leads  and  two  for  the  voltmeter  leads.  The 
utmost  care  must  always  be  exercised  in  using  the  proper 
terminals.  An  error  in  this  .respect  will  ruin  a  very  ex- 
pensive instrument. 


POWER  MEASUREMENT 


79 


FIG.  94. — Wattmeter 


Fig.  94,  shows  the  correct  method  of  connecting  a  watt- 
meter in  the  circuit  to  measure  the  power  taken  by  the  arc 
lamp  L.      The  ammeter  terminals  A  A, 
are  put  in  series  with  the    arc  lamp  L 
and    the    voltmeter    terminals    VV   are 
connected    in    parallel    across    the    arc 
lamp.     For  a  detailed  discussion  of  the 
wattmeter,  see  Chapter  XIV. 

64.  Work  —  Commercial    Units.      In 
order  to  compute   the  amount  of  work 
done  by  a  given  engine,  it  is  necessary 
to  know  the  time  it  has    been   running, 
and  the  power  it  has  been  supplying,  that 
is,  its  rate  of  doing  work.     If  the  power 
is  measured  in  HORSE-POWER  and  the  time  in  HOURS,  the 
work  done  is  measured  in  HORSE-POWER-HOURS  and  is  the 
product  of  the  Horse-power  by  the  Hours. 

Similarly,  if  the  power  is  measured  in  Kilowatts  and  the 
time  in  Hours,  the  work  done  is  measured  in  KILOWATT- 
HOURS  and  is  merely  the  product  of  the  Kilowatts  by  the 
Hours. 

The  HORSE-POWER-HOUR  and  the  KILOWATT-HOUR  are 
the  commercial  units  of  work. 

1  H.P.  hour  =.746  K.W.  hour; 
1  K.W.  hour  =  1.34  H.P.  hours. 

The  use  of  these  units  is  illustrated  by  the  following 
examples: 

Example.  How  much  work  is  done  in  one  day  of  8  hours  by 
a  150-K.W.  generator  running  at  full  load? 


or 


150X8-1200  K.W.  hours; 
1200X1.34  =  1610  H.P.  hours. 


Example.     At  15  cts.  per  K.W.  hour,  what  is  the  cost  of  burn- 
ing 100  lamps  for  8  hours  if  each  lamp  consumes  50  watts? 


80  ELEMENTS  OF  ELECTRICITY 

Power  consumed  in  100  lamps: 

100X50  =  5000  watts  =  5K.W.; 

8X5  =  40K.W.  hours; 
40X15  cts.  =  $6.00. 

65.  Work:  Small  Units— Foot-Pound :  Watt-Second  or 
Joule.  The  commercial  units  of  WORK,  the  H.P.  hour 
and  the  K.W.  hour,  are  too  large  to  be  used  conveniently 
in  some  problems.  Accordingly,  use  is  made  of  two  smaller 
units.  The  mechanical  unit  so  used  is  the  FOOT-POUND. 
The  electrical  unit  is  the  WATT-SECOND,  also  called  the 
JOULE. 

1  H.P.  hour  =1,980,000  ft.lbs. 

1  K.W.  hour  =3,600,000  watt-sees,  or  joules. 

1  Joule  =  .74  ft.lb. 

In  the  following  pages,  when  dealing  with  small  quantities 
of  work  the  foot-pound  and  watt-second  will  be  used ;  other- 
wise the  commercial  units  the  Horse-power-hour  and  the 
Kilowatt-hour  will  be  employed. 

Example.  How  much  work  is  done  when  a  25-watt  lamp  is 
burned  for  2  minutes? 

25  X  2  X  60  =  3000  watt-sees,  or  joules ; 
or 

3000 X. 74  =  2220  ft.lbs. 

The  following  are  the  electrical  units  of  WORK  and  POWER  in 
general  use: 

Work:  Watt-sec.  =  joule 
K.W.  hour; 

Power:  Watt 

Kilowatt 

Kilo  volt-ampere  (A.  C.  unit). 

WORK  =POWERXTIME. 

WORK 

POWER  =  — . 

TIME 


POWER  MEASUREMENT  81 

The  inter-relation  among  the  different  units  can  readily  be  seen 
by  the  following  table : 

Work  Units:  Watt-sec.    =joule  =  volt-coulomb; 

K.W.  hour  =  3,600,000  watt-sees. 
Power  Units:    Watt = joule  per  sec.  =  volt-ampere  =  volt-coulomb 

per  sec. 
Kilowatt  =  1000  watts. 

K. V. A.  (A.C.  unit)  =     Kllowatts     .     (See  chapter  XV 

Jrower  lactor 
and  tables  in  Appendix.) 

Problem  11-4.  How  much  will  it  cost  at  5  cts.  per  K.W.  hour 
to  run  a  110- volt  motor  for  8  hours?  Motor  takes  40  amperes. 

Problem  12-4.  If  work  in  Problem  11-4  we're  paid  for  at  the 
rate  of  4  cts.  per  H.P.  hour,  how  much  would  it  cost? 

Problem  13-4.  How  much  work  is  done  when  a  50-watt  lamp 
is  burned  for  4  hours? 

Problem  14-4.  At  12  cts.  per  K.W.  hour,  what  is  the  cost 
of  running  iron  in  Problem  5-4  for  6  hours? 

Problem  15-4.  What  work  is  done  in  maintaining  for  12  hours 
a  current  of  100  amperes  in  a  wire  of  .8  ohm  resistance? 

Problem  16-4.  A  bill  for  electric  energy  was  $16.40  for 
120  hours.  If  the  price  was  10  cts.  per  K.W.-hour,  what  was  the 
average  power  used? 

Problem  17-4.  When  100  incandescent  lamps  had  been  burned 
on  a  110-volt  circuit  for  4  hours,  a  bill  of  $3.00  was  presented, 
computed  on  the  rate  of  15  cts.  per  KM.  hour.  What  was  the 
average  current  taken  by  each  lamp? 

Problem  18-4.  At  5  cts.  per  K.W.  hour,  how  much  does  it 
cost  per  year  for  transmission  losses  in  Problem  52-3?  Lamps 
are  burned  2  hours  each  day. 

Problem  19-4.  What  is  the  total  energy  lost  in  the  line  wires 
per  month  of  100  hours  in  Problem  43-3? 

Problem  20-4.  By  how. many  foot-lbs.  is  1,000,000  watt-sees, 
greater  or  less  than  3  H.P.  hours? 

66.  Electrical  Energy  and  Heat  Energy.  An  electric 
current  may  be  used  in  one  part  of  a  circuit  to  produce 
mechanical  motion,  as  in  a  motor;  in  another  part  of  the 
circuit,  to  produce  electrolytic  action,  as  in  an  electroplat- 


82 


ELEMENTS  OF  ELECTRICITY 


ing  vat;  in  another  part,  to  produce  light,  as  in  an  electrio 
lamp.  In  any  part  of  the  circuit  where  it  is  doing  no  one 
of  these  things,  all  the  energy  consumed  goes  into  produc- 
ing heat.  It  even  produces  heat  in  the  portions  of  the 
circuit  where  it  is  also  producing  some  other  form  of  energy. 


FIG.  95.     Interior  of  transformer  showing  cooling  coils. 

A  motor  never  gives  out  in  mechanical  energy  all  that  it 
receives  in  the  form  of  electrical  energy.  Some  of  the 
electrical  energy  is  turned  into  heat  energy.  This  heat  is 
produced  in  overcoming  the  electrical  resistance,  just  as  heat 
is  produced  in  a  machine  in  overcoming  mechanical  friction. 


POWER  MEASUREMENT  83 

It  is  safe  to  say  that,  in  any  part  of  an  electric  circuit 
where  there  is  no  transformation  to  other  forms  of  energy,  the 
whole  of  the  electrical  energy  consumed  is  turned  into  heat. 

Thus  in  the  line  wires  in  Fig.  96,  all  the  electric  energy 
consumed  in  forcing  a  current  through  them  goes  into  heat.- 

The  resistance  of  an  electric  circuit  is  similar  to  the 
friction  of  a  machine.  Just  as  the  power  used  to  overcome 
the  resistance  of  the  wire  appears  in  the  heat  generated 
in  the  wire,  so  the  power  used  to  overcome  the  friction  of 
a  bearing  appears  in  the  heat  generated  in  the  bearing. 
The  mechanical  engineer  strives  to  reduce  the  amount  of 
power  wasted  in  heat,  by  reducing  the  friction  of  the  bear- 
ings. The  electrical  engineer  may  reduce  the  power  wasted 
in  heat,  by  reducing  the  resistance  of  the  wire  used  to  trans- 
mit a  given  current.  In  designing  electric  machinery,  a 
great  deal  of  care  is  taken  to  provide  sufficient  surface  for 
radiation,  so  that  the  heat  generated  may  not  cause  an 
excessive  rise  in  temperature. 

In  Fig.  95  the  interior  of  a  transformer  is  shown. 
Notice  the  coil  of  pipes  at  the  top.  Water  is  circulated 
through  these  pipes  in  order  to  carry  off  the  heat  gene- 
rated in  the  core  and  windings. 

Example.      The  generator  G  in  Fig.  96,  maintains  a  pressure 
of  240  volts  across  the  line.     The  line  wires  have  a  total  resistance 
of  4  ohms.      The   motor  M  re- 
quires 5   amperes   at  220   volts.  T     ~?  2ohms 
How  much  electrical   energy  is   -3  r^\ 
lost  in  heat  in  the  line  wire?           <=>  \^_J 

Solution.     Since  the  line  wire    *    '/ 2Ohma 

is   producing   no  other   form   of  FIG.  96. 

energy,  all   the   electrical  energy 

consumed  in  it  must  be  used  in  overcoming   its   resistance    and 

must   go    into   heating   it.     The    energy  used  in  the  line  wire  is 

found  as  follows: 

7  =  5  amperes; 

R  =  4  ohms. 

p  =  PR  =  5  x  5  X  4  =  100  watts  or  joules  per  sec. 


84  ELEMENTS  OF  ELECTRICITY 

Thus  100  joules  are  turned  into  heat  each  second.  If  this  heat 
is  not  allowed  to  radiate,  the  wire  would  soon  become  hot  enough 
to  destroy  the  insulation  and  finally  fuse  the  copper  itself.  Thus 
is  seen  the  necessity  for  arranging  wires  in  the  coils  of  a  trans- 
former and  in  the  field  and  armature  coils  of  motors  in  such  a  way 
'as  to  allow  for  the  dissipating  of  the  heat  generated  in  overcoming 
the  resistance  of  the  coils. 

Referring  again  to  Fig.  96,  it  is  seen  that  the  motor  consumes 
220X5  or  1100  watts.  The  greater  part  of  this  energy  is  used 
in  doing  the  mechanical  work  of  turning  the  armature.  But  a 
considerable  share  of  this  energy  even,  is  used  in  overcoming  the 
electrical  resistance  of  the  motor. 

Assuming  the  resistance  of  the  motor  is  2  ohms,  the  power 
consumed  in  overcoming  its  resistance,  and  thus  in  heating  it 
would  be  : 

P  =  PR; 
=  5X5X2  =  50  watts. 

Of  the  total  1100  watts  taken  by  the  motor,  not  more  than 
1050  watts  are  turned  into  mechanical  energy,  50  watts  being 
used  to  overcome  the  electrical  resistance  of  the  motor.  These 
50  watts  are  called  the  PR  Loss  or  sometimes  the  COPPER  Loss, 
since  the  electrical  conductors  are  generally  of  copper. 

67.  Heat  Equivalent  of  Electricity.  If  the  heat  gen- 
erated by  passing  an  electric  current  through  a  wire  be 
measured  by  allowing  it  to  raise  the  temperature  of  a  known 
weight  of  water,  it  is  found  that  one  watt-sec,  of  electrical 
energy  is  the  equivalent  of  .24  calorie.  That  is,  one  ampere 
flowing  through  an  ohm  resistance  for  1  sec.  will  raise  1  gm. 
of  water  .24°  C.  This  is  generally  stated  in  the  form  of  an 
equation  : 


where  #=Heat  in  calories; 

1=  Current  in  amperes; 
R=  Resistance  in  ohms; 
t  =time  in  seconds. 

Since 

I2R  =watts, 


POWER  MEASUREMENT  85 

then 

I2Rt  =  watt-sees,  or  joules; 

H  =.24  watt-sec,  or  joule. 

The  factor  (.24)  is  called  the  HEAT  EQUIVALENT  of  ELEC- 
TRICITY. 

Example.  Referring  again  to  Fig.  96,  we  find  that  100  watts 
were  expended  in  heating  the  lead  wires.  The  actual  number  of 
calories  thus  produced  per  sec.  would  be  100 X. 24  =  24  calories 
per  sec.  This  is  a  very  high  amount  of  energy  to  be  lost  in  trans- 
mitting such  a  small  quantity  of  electrical  energy  as  the  motor 
M  is  taking,  and  could  easily  be  avoided  by  using  line  wire  of  less 
resistance. 

Example.  A  bank  of  thirty- two  c.p.  incandescent  lamps  takes 
10  amperes  at  110  volts.  If  88  per  cent  of  the  energy  received 
by  the  lamps  is  given  off  in  heat,  how  many  calories  are  thus  given 
off  in  1  hour? 

10X110  =  1100  watts  (delivered  to  lamps); 
88%  of  1100  =  968  watts  (given  off  in  heat); 

1  watt  is  equivalent  to  .24  cal.  per  sec.; 
968  watts  =  968 X. 24  =  232  cal.  per  sec. 

232X60  =  13,900  cal.  per  min. 
13,900X60=835,000  cal.  per  hour. 

Or,  using  the  equation, 

//  =  .24PRt  =  .24  watt-sec. ; 
=  .24X(10X  110)  X3600X  (.88)  =835,000  cal. 

Problem  21-4.  How  much  heat  is  generated  per  hour  in  an 
electric  iron  using  3.5  amperes  at  110  volts? 

Problem  22-4.  If  the  price  of  the  electric  energy  used  in  Problem 
21-4  is  12  cts.  per  K.W.  hour,  what  is  the  price  per  calorie? 

Problem  23-4.  How  many  calories  per  hour  are  given  off  by  a 
car  heater  which  takes  6  amperes  at  550  volts? 

Problem  24-4.  How  much  heat  is  generated  in  10  hours  in 
the  lead  wires  of  Problem  18-3? 

Problem  25-4.  If  only  14  per  cent  of  the  electric  energy  used 
in  the  lamp  in  Problem  18-3  goes  into  light,  how  many  calories  of 
heat  does  the  lamp  give  off  in  burning  3  hours? 


86  ELEMENTS  OF  ELECTRICITY 

Problem  26-4.  How  long  would  a  current  of  5  amperes  under 
a  pressure  of  40  volts  have  to  run  to  supply  enough  heat  to  just 
melt  8  Ibs.  of  ice? 

68.  Efficiency.  Since  all  pieces  of  electrical  apparatus 
have  resistance,  there  must  always  be  some  heat  generated 
when  a  current  is  sent  through  them.  Unless  the  apparatus 
is  to  be  used  for  heating  purposes,  this  energy  is  wasted, 
just  as  the  energy  is  wasted  which  goes  toward  overcoming 
mechanical  friction.  It  follows  from  this  that  no  electrical 
machine  is  perfect.  That  is,  it  does  not  give  out  in  useful 
energy,  all  the  energy  that  is  put  into  it.  We  can  say,  then, 
that  no  electrical  machine  has  100  per  cent  efficiency. 

When  we  speak  of  the  commercial  efficiency  of  electrical 
machinery  we  mean  the  same  that  we  do  in  speaking  of  the 
efficiency  of  any  machine;  the  percentage  that  the  useful 
output  is  of  the  input. 

In  many  commercial  tests,  it  is  difficult  to  measure 
directly  the  input  into  a  machine.  But  it  may  be  com- 
puted from  the  output  and  the  losses,  since  the  input  must 
equal  the  output  plus  the  losses. 

We  then  have  the  equation: 

output  output 

Efficiency  =-r—    -r  =  —  -,  -  . 

input      output  -f  losses 

Example.     A   3-H.P.   motor   requires   2.4   K.W.   to  drive   it. 
hat  is  its  efficienc? 


What  is  its  efficiency 


Output  =  3.0  H.P.; 
Input    =2.4X1.34=3.2  H.P.; 
Output     3.0 


For  a  discussion  of  the  efficiency  of  motors  and  generators  see 
Chapter  IX. 

Problem  27-4.     What  efficiency  has  a   15-H.P.  motor  which 
requires  65  amperes  at  220  volts? 


POWER   MEASUREMENT  87 

Problem  28-4.  A  generator  has  an  efficiency  of  90  per  cent. 
What  current  can  it  deliver  at  a  pressure  of  110  volts  when  it 
receives  10  H.P.  from  the  driving  engine? 

Problem  29-4.  What  is  the  efficiency  of  a  motor  which  does 
428,000  ft.lbs.  of  work  in  5  minutes,  if  the  input  is  2.5  K.W.? 

69.  Efficiency  of  Transmission.  When  energy  is  trans- 
mitted from  one  place  to  another,  some  energy  is  always 
used  in  the  transmission  line.  In  mechanical  transmission 
of  energy,  there  is  the  pulley  and  belt  friction,  etc.,  to  be 
overcome.  In  electrical  transmission,  the  resistance  of  the 
lead  wires  must  be  overcome.  That  method  of  transmission 
is  most  efficient  which  loses  the  smallest  percentage  of 
energy  in  the  line.  The  efficiency  of  transmission  may 
be  defined  as  that  percent- 
age which  the  useful  output  J  ~f  *  "^  ohms 
of  the  line  is  of  the  total  in- 

PUt'  !•      X  .2  Ohms 

Example.     The  lamp  bank  A,  F'«-  97. 

Fig.  97,  uses  12  amperes.     What 

is  the  efficiency  of  the  transmission  ;   that  is,  what  per  cent  of  the 
total  power  delivered  to  the  line  is  used  by  the  lamps? 

Total  power  =  (IE)  =  120  X  12  =  1440  watts. 

Watts  lost  in  line  =  (PR)  =  12  X  12  X  .4  =  57.6  watts. 

Watts  left  for  lamps  =  1440-  57.6  =  1382.4  watts. 

1382 
Efficiency  of  transmission  =  —  —  =  96%; 


or 

Volts  lost  in  line  =  (IK)  =  12  X  .4  =  4.8  volts. 
Volts  across  lamps  =  120  -4.8  =  115.2  volts. 
Watts  delivered  to  lamps  =  (IE)  =  1  15.2  X  12  =  1382  watts. 


Problem  30-4.  What  is  the  efficiency  of  transmission  for  a 
circuit  which  receives  4  K.W.  from  a  generator  at  one  end  and 
delivers  3.8  K.W.  to  a  motor  at  the  other  end? 


88  ELEMENTS  OF  ELECTRICITY 

Problem  31-4.    What  is  the  efficiency  of  transmission  in  Prob- 
lem 42-3? 

Problem  32-4.     What  is  the  efficiency  of  transmission  in  Prob- 
lem 34-3? 

Problem  33-4.     What  is  the  efficiency  of  transmission  in  Prob- 
lem 38-3? 

Problem  34-4.     What  is  the  efficiency  of  transmission  in  Prob- 
lem 51-3? 

Problem  35-4.     What  is  the  efficiency  of  transmission  in  Prob- 
lem 43-3? 

Problem  36-4.     What  is  the  efficiency  of  transmission  in  Prob- 
lem 52-3? 

Problem  37-4.     What  is  the  efficiency  of  transmission  in  Prob- 
lem 53-3? 


70.  Efficiency  of  Lamps.  Regarding  a  lamp  as  a 
machine,  we  may  define  the  efficiency,  as  the  ratio  of  the 
watts  given  out  in  light,  to  the  total  watts  received  at  the 
lamp  terminals.  This  is  a  very  instructive  way  to  study 
the  performance  of  a  lamp.  The  A.I.E.E.  in  1907  denned 
the  efficiency  of  an  incandescent  lamp  as  the  candle-power 
produced  by  each  watt  received  by  the  lamp.  Thus  a 
lamp  taking  .5  ampere  at  110  volts,  or  55  watts,  and  giving 

14 

14  c.p.,  would  be  rated  according  to  the  A.I.E.E.  as  — -  or 

55 

.25  c.p.  per  watt. 

Neither  of  these  methods,  however,  is  in  common  use 
in  giving  the  efficiency  of  any  kind  of  electric  light.  The 
general  practice  is  to  state  the  number  of  watts  required 
to  produce  1  c.p.  of  light.  The  above  55-watt,  14  c.p.  lamp 

would  then  be  rated  as  a  lamp  of  —  or  3.93  watts  per  c.p. 

Thus,  the  fewer  the  watts  used  per  c.p.,  the  more  efficient 
the  lamp.  An  arc  lamp  using  .8  watt  per  c.p.  is,  therefore, 
much  more  efficient  than  an  incandescent  lamp  using 
3.93  watts  per  c.p.  as  above. 


POWER   MEASUREMENT  89 

The  problem  of  illumination  is  as  much  a  matter  of  even 
distribution  of  the  light  as  of  the  high  efficiency  of  the 
lamp.  The  ideal  lamp,  of  course,  combines  a  high  efficiency 
with  an  even  distribution.  For  a  fuller  discussion  of  the 
subject  of  illumination,  see  Chapter  XIII. 

Problem  38-4.  A  carbon  filament  lamp  of  16  c. p.  takes  55  watts. 
What  is  the  light  efficiency  of  the  lamp? 

Problem  39-4.  A  tungsten  lamp  is  rated  as  a  50-watt  lamp. 
If  the  light  efficiency  is  1.5  watts  per  c.p.,  how  many  candle-power 
has  it? 

Problem  40-4.  A  certain  32  c.p.  incandescent  lamp  gives 
out  13  watts  in  light  and  89  watts  in  heat,  (a)  What  is  the  effi- 
ciency of  the  lamp  as  a  machine  for  a  source  of  heat?  (6)  As 
a  source  of  light?  (c)  What  would  be  the  commercial  rating 
of  the  lamp? 


90  ELEMENTS  OF  ELECTRICITY 


SUMMARY    OF    CHAPTER   IV 

USE  OF  AMMETER.  Measures  current;  has  low  resist- 
ance; always  put  in  series  with  that  part  of  the  circuit  in 
which  the  current  is  to  be  measured. 

USE  OF  VOLTMETER.  Measures  voltage;  has  high 
resistance;  connected  in  parallel  across  the  points,  between 
which  the  voltage  is  to  be  measured. 

USE  OF  WATTMETER.  Measures  power;  is  a  combina- 
tion of  ammeter  and  voltmeter.  Ammeter  side  is  connected 
in  series  with  circuit,  and  voltmeter  side  across  circuit.  Great 
care  must  be  used  not  to  get  ammeter  side  across  the  circuit. 

COMPUTATION  OF  ELECTRICAL  POWER. 
Watts  (power)  =  amperes  X  volts. 

This  should  be  understood  to  mean  that  the  Watts  consumed 
in  any  part  of  a  circuit  equals  the  product  of  the  Amperes 
flowing  through  that  same  part  of  the  circuit  times  the  Volts 
across  that  same  part.  Just  as  Ohm's  law  may  be  written  in 
three  ways,  so  the  power  equation  may  be  written  in  three 
ways,  thus: 


HEAT  EQUIVALENT  OF  ELECTRICAL  ENERGY. 

i  watt-sec,  (joule)  =  .24  calorie. 

The  total  amount  of  heat  developed  by  an  electric  current 
flowing  for  a  given  time  through  a  resistance  is  computed  from 
the  following  equation: 

H  =  .24l2Rt, 

where  H  is  measured  in  calories  ; 
I  "          amperes; 

R  "          ohms; 

t  "          seconds. 

.24  is  the  Electrical  Heat  Equivalent. 

EFFICIENCY  OF  ELECTRIC  MACHINES.  The  com- 
mercial efficiency  of  an  electric  machine,  as  of  any  machine, 
may  be  stated  as  the  percentage  of  useful  energy  given  out  by 
the  machine,  in  comparison  with  the  energy  put  into  the 
machine.  The  same  is  true  of  an  electric  transmission. 


POWER   MEASUREMENT  91 

Electric  lamps  may  be  regarded  as  machines  and  their 
efficiency  stated  as  above.  The  usual  method,  however,  is  to 
state  the  number  of  watts  consumed  per  candle-power  of  light 
given  out. 

The  efficiency  of  generators  is  generally  most  easily  com- 
puted from  the  equation: 

output 


Commerical  efficiency 


output  +  losses" 


PROBLEMS   ON   CHAPTER  IV     . 

41-4.  The  Western  D.  C.  Ammeters  are  designed  to  have  45 
millivolts  drop  across  them  when  they  are  giving  their  full  scale 
reading. 

(a)  What  must  be  the  resistance  of  a  1  ampere  ammeter? 

(b)  Of  a  50  ampere  ammeter? 

(c)  Of  a  15  ampere  ammeter? 

(d)  How  much  current  would  15  ampere  ammeter  take  if  placed 
by  mistake  across  110  volts? 

42-4.  A  Weston  D.C.  Voltmeter  has  approximately  100  ohms 
for  each  volt  of  the  scale.  What  current  does  a  150-volt  volt- 
meter take  when  placed  across  a  110  volt  circuit? 

43-4.  Motor  in  Fig.  98  takes  , 

20  amperes  at  110  volts.  /~          " Ohms 

(a)  What  is   the   "  brush   po-  S^\                                PM! 

tential  "  of  the  generator  Gl  ^-/ 

(6)  What  is   the   efficiency   of        / 1.5  ohms 


transmission?  FlG  98 

44-4.  An  engine  supplies  160 

H.P.  to  a  generator  delivering  200  amperes  at  550  volts.     What 
is  the  commercial  efficiency  of  the  generator? 

45-4.  A  generator  delivers  a  current  of  90  amperes  at  a  pressure 
of  110  volts.  What  power  does  it  supply  in  kilowatts?  How 
many  H.P.? 

46-4.  The  generator  in  Problem  45  has  a  shunt-wound  field  with 
a  resistance  of  200  ohms.  How  many  watts  are  absorbed  in  the 
field? 


92  ELEMENTS  OF  ELECTRICITY 

47-4.  The  armature  of  generator  in  Problem  45  has  a  resistance 
of  .02  ohm.  What  power  is  lost  in  the  armature  conductors? 

48-4.  An  engine  supplies  150  H.P.  to  a  generator  delivering 
180  amperes  at  a  pressure  of  550  volts.  What  is  the  commercial 
efficiency  of  the  generator? 

49-4.  Assuming  an  efficiency  of  50  per  cent  for  the  whole 
arrangement,  what  current  will  be  used  by  a  250-volt  electric 
hoist  when  raising  2500  Ibs.  200  ft.  per  minute? 

50-4.  Assuming  that  70  per  cent  of  the  electrical  input  is  use- 
fully expended,  what  current  will  be  required  by  a  500-volt  rail- 
way motor  in  propelling  a  60-ton  car  up  a  2  per  cent  grade  at  a 
speed  of  8  miles  per  hour? 

61-4.  Electrical  energy  for  lighting  costs  15  cts.  per  K.W.  hour, 
what  is  the  cost  per  month  (30  days)  of  operating  a  16  c.p. 
lamp,  which  takes  3.1  watts  per  candle,  the  lamp  being  in  service 
3  hours  each  day? 

52-4.  Electrical  energy  is  supplied  at  5  cts.  per  K.W.  hour 
for  driving  a  10-H.P.  motor.  The  efficiency  of  the  motor  at  full 
load  is  85  per  cent.  Find  the  cost  of  operating  the  motor  for 
100  hours. 

53-4.  A  current  of  0.5  ampere  flowing  through  a  lamp  generates 
180  calories  of  heat  in  12  seconds,  (a)  What  is  the  resistance 
of  the  lamp?  (b)  What  power  is  expended  in  the  lamp?  Express 
in  watts  and  in  H.P. 

54-4.  A  field  coil  of  a  generator  contains  20  Ib.  of  copper  (specific 
heat  0.095);  weight  of  insulation  negligible.  The  resistance  of 
the  coil  is  200  ohms.  How  fast  does  the  temperature  of  the  coil 
rise  when  a  current  of  2.1  ampere  is  flowing  through  the  coil? 
Allow  for  no  radiation. 

55-4.  How  long  would  it  take  for  the  temperature  of  the  coil 

in  Problem  54-4  to  rise  50°  C., 

9  -so*"*  P  .4 ohm  E  if  no  heat  were  given  off  from 

/\  /]\B  the  coil  in  radiation? 

\/          AT/  66"*'  Each  lamp'   FiS-"> 

.3  ohm  \^/   .4Ohm\|/         takes  1  ampere. 

M  -     /     G  F  Find: 

Fl(5  ^  (a)  Voltage  across  group  A 

and  across  B. 

(b)  Volts  lost  in  CD  and  GM  and  in  DE  and  FG. 

(c)  Watts  delivered  to  each  group. 

(d)  Watts  lost  in  line.      **-  ' 

(e)  Efficiency  of  transmission. 


POWER  MEASUREMENT  93 

57-4.  Voltage  across  terminals  of  arc  lamp  in  Fig.  100  is 
110  volts.  If  arc  takes  5  amperes  at  85  volts  and  gives  800  c.p. 

(a)  What  power  is  consumed  by  the  series  resist- 
ance 72? 

(b)  What  is  efficiency  of  the  lamp,  as  a  machine? 
as  a  lamp? 

58-4.  At  8  cts.  per  K.W.  hour,  how  much  will  it 
cost,  per  week  of  60  hours,  to  run  a  motor,  having 
an  average  load  of  40  H.P.,  and  an  average 
efficiency  of  90  per  cent? 

59-4.   What    must  be  the  H.P.  of    an  engine  to 
run  a  generator  feeding  600  ^-ampere  lamps  at  110 
volts?     Five  volts  are  lost  in  the  line,  the  efficiency 
of  the  generator  is  92  per  cent,  and  the  loss  in   the  belt   is  1  per 
cent. 

60-4.  At  Group  II,  Fig.  101,  there  are  four  55-watt  110-volt 
lamps.  Group  I  consists  of  five  80- watt  11 5- volt  lamps.  Internal 
resistance  of  generator  G  is  4  ohms. 

Find: 

(1)  Resistance  of  AC  +  BD. 

(2)  Volts  lost  between  G  and  Group  I. 

(3)  Brush  potential  of  generator. 

(4)  E.M.F.  of  generator. 

(5)  Efficiency  of  transmission. 

(6)  Average  resistance  per  lamp  of  Group  I. 

(7)  Average  resistance  per  lamp  of  Group  II. 

1  Ohm       A  C 

In 


6666 


1  Ohm 


B  D 

FIG.  101. 

61-4.  Generator,   Fig.   102,   delivers  5  K.W.  to  line.      Motor 
uses  4.8  K.W. 
Find: 

(a)  Watts  lost  in  line. 

(b)  Voltage  and  current  of  motor. 

(c)  Brush  potential  of  generator. 


94 


ELEMENTS  OF  ELECTRICITY 


62-4.  Assume  internal  resistance  of  generator  in  Problem  61 
to  be  4  ohms.  The  owner  of  the  motor  pays  the  owner  of  the 
generator  10  cts.  per  K.W.  hour  for  energy.  If  it  costs  the  owner 
of  the  generator  9  cts.  per  K.W.  hour  to  generate  energy,  how 
much  does  he  earn  or  lose  in  a  month  of  250  hours? 


2  Ohms 


lOhm 


H 


FIG.  102. 

63-4.  Resistance  of  shunt  coil  S  in  arc  lamp  in  Fig.  103  is 
440  ohms.  Voltage  across  terminals  =  55  volts.  Arc  A  requires 
7  amperes  at  35  volts  arid  gives  600  c.p. 

Find: 

(a)  Resistance  of  R. 

(b)  Power  consumed  by  lamp. 

Power  consumed  by  resistances  S  and  R. 
Efficiency  of  lamp, 

(1)  As  a  machine ; 

(2)  As  a  lamp. 

64-4.  A  wire  having  a  resistance  of  40  ohms  is  coiled  in  a 
vessel  containing  300  gin.  of  oil  of  which  the  specific  heat  is  0.75. 
The  vessel  is  copper  and,  together  with  copper  wire,  weighs  200  gm. 
(Specific  heat=  .095.)  What  will  be  the  temperature  rise  after  a 
current  of  4.2  amperes  has  passed  through  the  coil  of  wire  for  5 
minutes?  Neglect  radiation. 


(d) 


CHAPTER  V 
MEASUREMENT  OF  RESISTANCE 

Circular  Measure — Mil — Mil-foot — Resistivity — Temperature  Co- 
efficient of  Resistance — Use  of  Wire  Tables — Temperature 
measured  by  Change  in  Resistance — Methods  of  Measuring  Re- 
sistance; Fall  of  Potential,  Ammeter-voltmeter,  Wheatstone 
Bridge,  Voltmeter  Method — Insulation  Resistance  by  Galvan- 
ometer Deflection  and  Standard  Megohm — Location  of  Faults. 

71.  Resistance — A  Means  of  Current  Control.  We  have 
seen  in  the  previous  work  that  the  amount  of  current  flow- 
ing in  a  given  circuit  depends  upon  the  resistance  of  the 
circuit  and  the  voltage  across  it.  This  is  also  true  of  any 
Part  of  a  circuit.  We  can,  therefore,  control  the  amount 
of  current  either  by  controlling  the  voltage  or  the  resistance, 
or  both.  The  control  of  the  resistance  is  generally  the 
easiest,  and,  in  many  cases,  the  only  available  method. 
It  is  therefore  necessary  to  become  familiar  with  the  com- 
mon methods  of  computing  and  regulating  the  resistance 
of  the  several  parts  of  a  circuit. 

For  example,  in  line  or  "  feed  "  wires  of  a  system,  there 
should  be  as  little  resistance  to  the  flow  of  the  current  as  is 
practicable.  For  this  reason  copper  is  generally  used  when 
much  power  is  to  be  transmitted.  The  cost  of  copper  is 
high,  so  we  do  not  wish  to  use  any  more  than  is  necessary. 
The  wires  then  should  be  as  small  as  is  practicable.  We 
must  make  the  wire  large  enough  to  transmit  the  current, 
but  small  enough  to  avoid  excessive  cost  of  installation. 
Thus,  it  is  necessary,  even  before  it  is  put  up,  to  be  able  to 
compute  the  resistance  of  a  given  length  of  copper  wire 
of  a  certain  cross-section  area  or  weight. 

95 


96  ELEMENTS  OF  ELECTRICITY 

72.  Circular-Mil  Measure.  Inasmuch  as  most  wire  is 
drawn  round,  it  is  inconvenient  always  to  retain  the  old 
method  of  measuring  the  cross-sectional  area  in  square 
inches.  The  cross-section  is  circular,  not  square;  so  we  use 
a  circular  measure,  and  not  a  square  measure.  In  circular 
measure  we  make  use  of  a  circular  unit  of  area,  instead  of 
a  square  unit  of  area.  This  circular  unit  of  area  is  a  CIRCU- 
LAR MIL;  .  symbol,  C.M.  or  cir.  mil.  It  is  the  area  of  a 
circle  whose  diameter  is  1  mil  in  length. 

The  term  "  mil  "  always  means  7™:.     As  in  our  coinage 


1  mill  =  n  nnn  of   a  dollar,  so  in  distance,  1  mil=—  —  of  1 
J.UUU  J.UUU 

inch. 

One  CIRCULAR  MIL,  then,  is  the  area  of  a  circle,  the  diam- 
eter of  which  is  one  mil  or  one-thousandth  of  an  inch.  Just 
as  a  square  mil  is  the  area  of  a  square  whose  sides  are  1 
mil  or  one-thousandth  of  an  inch. 

The  area  of  a  square  whose  sides  measure  5  mils  is  5X5 
or  25  sq.mils.  Similarly  the  area  of  a  circle  whose  diameter 
is  5  mils  is  5X5  or  25  circular  mils.  To  find  the  number  of 
circular  mils  in  a  circle  of  a  given  diameter;  we  have  merely 
to  square  the  number  of  mils  in  the  diameter. 

Examples.     (1)  Find  area  in  circular  mils  of  a  circle  of   .015 
inches  in  diameter. 
.015  in.  =15  mils. 

Area  of  circle  =  15X15  =  225  cir.  mils.    Ans. 
(2)  Find  area  of  a  circle  of  .346  in.  dia. 
.364  in.  =364.0  mils. 

Area  of  circle  =  364  X  364  =  132,500  cir.  mils.     Ans. 
The  advantage  of  measuring  circular  sections  in  circular  measure 
is  too  obvious  to  be  mentioned. 

Problem  1-6.  Find  the  cross-section  area  m  C.M.  of  a  round 
wire  .164  in.  in  diameter. 

Problem  2-5.  What  is  the  circular  mil  area  of  a  circle  2,18  in, 
in  diameter? 


MEASUREMENT  OF  RESISTANCE  97 

Problem  3-5.  Find  the  cross-section  area  of  a  round  wire  ^g  in. 
in  diameter. 

Problem  4-5.  A  round  wire  has  a  cross-section  area  of  500,000 
C.M.  What  is  its  diameter  in  inches? 

Problem  5-5.  (a)  Find  the  area  of  the  circle  in  Problem  3  in 
square  inches.  (fr)  In  square  mils. 

73.  Computation  of  Resistance  from  Size,  etc.  ;  Mil-Foot. 

Just  as  in  hydraulics  the  resistance  of  a  large  pipe  is  less 
than  the  resistance  of  a  small  one,  so  the  electrical  re- 
sistance of  a  large,  wire  is  less  than  the  resistance  of  a  small 
wire.  Thus  if  the  resistance  of  a  wire  of  1  cir.mil  cross- 
section  were  20  ohms,  the  resistance  of  a  wire  of  4  cir.mils 
would  be  not  4  times  as  great,  but  J  as  great,  or  J  of  20  = 
5  ohms. 

Of  course  the  resistance  varies  directly  as  the  length; 
a  wire  3  times  as  long  as  another  of  same  cross-section 
would  have  3  times,  the  resistance.  We  take  for  our  unit 
wire,  1  MIL-FOOT.  A  wire  ONE  FOOT  long,  and  having  a 
cross-section  area  of  ONE  CIRCULAR  MIL  is  called  a  MIL- 
FOOT  wire.  In  order,  then,  to  find  the  resistance  of  a 
certain  number  of  feet  of  wire  of  given  cross-section,  we 
merely  have  to  multiply  the  resistance  of  a  mil-foot  by  the 
length  in  feet  and  divide  by  the  sectional  area  in  circular 
mils.  This  is  evident  from  the  fact  that  the  resistance  of 
a  wire  increases  as  the  length  increases,  but  decreases  as 
the  section  area  increases. 

This  is  generally  expressed  in  the  form  of  an  equation: 


where  R  =  resistance  of  wire  in  ohms; 

K=  "          1  mil-ft.  in  ohms; 

I  =  length  in  feet  ; 

d=diameter  in  mils; 
or 

d2  =section  area  in  cir.-mils, 


98  ELEMENTS  OF  ELECTRICITY 

This  resistance  per  mil-ft.  (K)  is  sometimes  called  the 
RESISTIVITY  or  SPECIFIC  RESISTANCE  of  a  material,  and 
varies  \Aith  different  materials.  See  Appendix. 

Example.  Assuming  the  resistance  per  mil-ft.  of  copper  at 
20°  C.  as  10.4  ohms,  what  is  the  resistance  of  500  ft.  of  copper  wire, 
.021  in.  in  diameter? 

7?     Kl. 
-ffi> 

10.4X500  ft. 


- 
(21X21)  cir.  mils. 


ohms. 


Example.  What  size  copper  wire  must  be  used  to  transmit  an 
electric  current  4  miles,  if  the  line  wire  resistance  is  not  to  exceed 
2  ohms? 


o_  10.4  X4X5280_  219000 


or 

d2  =  109,500  cir.mil. 
d  =  33l  mils.  diam. 
d=.331  in.  diam. 

NOTE:  Unless  otherwise  stated,  the  temperature  of  wires  in 
all  problems  is  to  be  taken  as  20°  C. 

Problem  6-5.  What  is  the  resistance  of  a  copper  wire  4000  ft. 
long  and  \  in.  in  diameter? 

Problem  7-5.  What  length  of  copper  wire  .064  in.  in  diameter 
will  have  a  resistance  of  8  ohms? 

Problem  8-5.  What  diameter  will  a  wire  one  mile  long  have, 
if  the  resistance  is  one  ohm? 

Problem  9-5.  It  is  desired  to  transmit  400  amperes  to  a  point 
2500  ft.  from  the  generator,  with  not  more  than  5  volts  "  line 
drop."  What  diameter  copper  wire  must  be  used? 


MEASUREMENT  OF  RESISTANCE  99 

Problem  10-6.  What  current  can  be  transmitted  over  a  copper 
wire  2  miles  long,  -fa  in.  in  diameter  with  but  8  volts  "  line  drop  "? 

Problem  11-5.  What  line  drop  is  there  in  a  4-mile  copper 
trolley  wire  carrying  200  amperes,  if  the  wire  is  .325  in.  in  diameter? 

74.  Resistivity  of  Metals  other  than  Copper.     Although 
copper,   on   account    of   its    low    resistivity,   is    the   metal 
most  widely  used  for  electrical  conductors,  aluminum  and 
even  galvanized  iron  are  sometimes  used.      The  resistivity 
of  aluminum  is  18.7  ohms  per  mil-ft.  at  20°  C.,  nearly  twice 
that  of  copper.     But  its  low  specific    gravity  more  than 
counterbalances  this,  so  that  for  equal  lengths  and  weights 
aluminum  wire  has  less  resistance  than  copper,  and  for  this 
reason  is  coming  into  more  general  use. 

The  resistivity  of  iron  and  steel  is  about  seven  times  that 
of  copper.  These  materials,  therefore,  can  be  used  only 
where  a  conductor  of  a  large  cross-section  can  be  installed, 
as  in  the  case  of  a  third  rail,  or  where  very  little  current 
is  to  be  transmitted,  as  in  the  case  of  the  telegraph.  For 
the  resistivity  of  various  pure  metals  and  alloys,  see 
Appendix. 

« 

Problem  12-5.  What  is  the  resistance  of  a  mile  of  aluminum 
wire,  .084  in.  in  diameter? 

Problem  13-5.  What  diameter  would  a  wire  of  aluminum 
have  to  be  to  fulfill  conditions  of  Problem  8-5? 

Problem  14-5.  If  a  steel  rail  were  used  in  Problem  9-5,  what 
cross-section  (sq.in.)  would  be  necessary? 

75.  Temperature  Coefficient  of  Resistance.     It   will   be 
noticed  that  when  the  resistance  of  a  mil-ft.  of  copper  wire 
was  given  as  10.4  ohms,  and  of  aluminum  as  18.7  ohms, 
that  the  metal  was  assumed  to  be  at  a  temperature  of  20° 
C.     The  reason  for  stating  the  temperature  is  that  the  re- 
sistance of  all  pure  metals  increases   as  the  temperature 
rises.     The  amount  that  the  resistance  increases  per  degree 


100  ELEMENTS  OF   ELECTRICITY 

rise  of  temperature  for  each  ohm  of  the  resistance  at  the 
standard  temperature  is  called  the  TEMPERATURE  COEF- 
FICIENT of  RESISTANCE.  For  all  pure  metals  this  coefficient 
is  nearly  the  same,  lying  between  the  values  .003  and  .006, 
and  depends  upon  what  is  taken  as  the  standard  or  initial 
temperature. 

Taking  the  resistance  at  0°  C.  as  the  standard,  copper 
increases  .0042  ohm  per  degree  for  each  ohm  at  0°.  Thus 
if  the  resistance  of  a  copper  wire  is  80  ohms  at  0°  C.  it  will 
be  increased  by  80  X.  00420  or  .336  ohm  for  each  degree 
the  temperature  rises  above  0°  C.  At  50°  C.  the  resistance 
would  have  increased  50  X.  336  or  16.8  ohms.  The  re- 
.sistance  then  at  50°  C.  would  be  the  resistance  at  0°  C.  plus 
the  increase  or,  80+16.8=96.8  ohms.  Other  metals  in- 
crease at  a  slightly  different  rate.  See  Appendix. 

The  resistance  of  a  mil-ft.  of  commercial  copper  at  0°  C. 
is  variously  given,  because  a  very  small  trace  of  any  im- 
purity raises  the  resistance  perceptibly.  According  to 
A.I.E.E.  standard  tables,  the  resistance  per  mil-ft.  at  0°  C.  = 
9.55  ohms.  According  to  G.  E.  tables,  the  resistance  per 
mil-ft.  at  0°  C.  =9.60  ohms. 

In  order  to  find  the  resistance  per  mil-ft.  at  any  other 
temperature,  it  is  merely  necessary  to  find  the  increase  in 
resistance  due  to  temperature  rise  and  add  this  increase 
to  the  resistance  at  0°  C.  Thus  the  resistance  per  mil-ft. 
at  25°  C.  =9.6  +  (9.6  X.  0042X25)  =10.6  ohms. 

This  effect  of  temperature  on  resistance  of  copper  may  be 
expressed  in  the  form  of  an  equation  : 


where  R  =  Resistance  at  temperature  above  or  below  0°  C. 
R0=         "          "  0°C.; 
T  =  Temperature  of  wire  in  degrees  Centigrade. 

Note:  When  using  the  temperature  coefficient  .0042  it  is  always 
necessary  to  use  the  Resistance  at  0°  C.  as  standard.  The  G.  E. 
Co.  has  established  as  a  secondary  standard  the  resistance  at  25°  C. 


MEASUREMENT  OF   RESISTANCE  ,*  ,  -,  :,    1QL 

The  coefficient  then  becomes  .00381.     And  the  equation  for  effect 
of  temperature  on  resistance  becomes  : 


where  R  =  Resistance  at  temperature  above  or  below  25°  ; 
R25=        "          "   25°  C.; 
T  =  Number  of  degrees  wire  is  above  or  below  25°  C. 

The  coefficient  .00381  must  be  used  because  the  standard  resist- 
ance at  25°  C.  would  be  larger  than  at  0°  C.  and  thus  the  increase 
per  degree  for  each  ohm  at  25°  C.  would  be  smaller  than  for  each 
ohm  at  0°  C.,  though  of  course  the  increase  for  the  total  resistance 
would  be  the  same. 


Example.     The  resistance  of  a  copper  wire  is  2.48  ohms  at 
25°  C.     What  will  the  resistance  of  this  wire  become  at  40°  C.? 


#40  =  #25(1  +  .  00381 T) 
=  2.48(1 +  .00381X15) 
=  2.48X1.0572 
=  2.62  ohms. 

Problem  16-5.  The  resistance  of  a  copper  wire  is  4.90  ohms  at 
0°  C.  What  is  it  at  20°  C.? 

Problem  16-5.  The  resistance  of  the  field  coils  of  a  generator 
is  220  ohms  at  25°  C.  When  the  coils  become  heated  to  75°  C. 
what  will  the  resistance  be? 

Problem  17-5.  What  will  the  resistance  of  a  coil  of  copper 
wire  become  at  7°  C.  if  the  resistance  is  200  ohms  at  25°  C.? 

Problem  18-5.  What  will  the  resistance  of  a  coil  of  copper 
wire  become  at  7°  C.,  if  the  resistance  is  200  ohms  at  0°  C.? 

Problem  19-5.  A  coil  has  a  resistance  of  12.5  ohms  at  0°  C. 
How  high  will  the  temperature  be  when  the  resistance  reaches  14.5 
ohms? 


76.  Coefficients  for  Various  Initial  Temperatures.  It  is  pos- 
sible to  regard  any  temperature  as  the  standard,  providing  the 
coefficient  for  that  standard  is  used.  There  is  thus  a  different 
coefficient  for  each  initial  temperature  that  is  used  as  a  standard. 
These  coefficients  are  given  by  the  AJ.E.E.  in  the  following  tables; 


102 


ELEMENTS  OF  ELECTRICITY 


TEMPERATURE  COEFFICIENTS  FOR  COPPER 

i  =  initial  or  standard  temperature, 

a  =  coefficient,  or  change  in  resistance  per  degree  C.  for  each  ohm 
at  temperature  Ti. 


Ti 

a 

Ti 

a 

Ti 

a 

0 

.00420 

17 

.00392 

34 

.00368 

1 

.00418 

18 

.00391 

35 

.00366 

2 

.00417 

19 

.00389 

36 

.00365 

3 

.00415 

20 

.00388 

37 

.00364 

4 

.00413 

21 

.00386 

38 

.00362 

5 

.00411 

22 

.00385 

39 

.00361 

6 

.00410 

23 

.00383 

40 

.00360 

7 

.00408 

24 

.00382 

41 

.00358 

8 

.00406 

25 

.00381 

42 

.00357 

9 

.00405 

26 

.00379 

43 

.00356 

10 

.00403 

27 

.  00377 

44 

.00355 

11 

.00402 

28 

.00376 

45 

.00353 

12 

.00400 

29 

.00374 

46 

.  00352 

13 

.00398 

30 

.00373 

47 

.00351 

14 

.00397 

31 

.  .00372 

48 

.00350 

15 

.00395 

32 

.00370 

49 

.00348 

16 

.00394 

33 

.00369 

50 

.  00347 

By  means  of  this  table  we  may  always  use  the  same  simple 
equation, 

Rt  =  Ri(l±aT); 

where  Ri  =  resistance  at  initial  temperature; 
Rt=      "  "  second 

T  =  change  of  temperature; 
a  =  coefficient  at  initial  temperature. 

Example.  The  resistance  of  a  coil  of  copper  wire  at  19°  C. 
is  240  ohms.  What  will  be  the  resistance  of  the  same  coil  at 
42°  C.? 


From  table,  "a"  at  19°  =  .00389; 

Rt  =  240(1  +  .00389X23) 
-240(1  +  .0895) 
=  261  ohms. 


Resistance  at  42°  C.  =  261  ohms. 


MEASUREMENT  OF   RESISTANCE  103 

Problem  20-5.  The  resistance  of  a  field  coil  is  130  ohms  at 
20°  C.  What  will  it  be  at  180°  C.? 

Problem  21-5.  What  will  be  the  resistance  of  a  copper  wire 
at  10°  C.  if  the  resistance  at  45°  C.  is  2.08  ohms? 

77.  Temperature  Measured  by  Resistance.  Electrical 
machines  are  generally  sold  under  a  guarantee  that  the 
wire  in  the  coils,  etc.,  will  not  rise  more  than  a  given  number 
of  degrees,  when  running  under  specified  load  for  a  specified 
time. 

By  measuring  the  resistance  of  the  coils  when  at  room 
temperature  and  then  again  at  close  of  run,  and  applying 
the  equation  for  temperature  effect,  the  temperature  rise 
can  easily  be  found. 

Example.  The  primary  coils  of  a  transformer  have  a  resistance 
of  5.48  ohms  at  25°  C.  After  a  run  of  2  hours,  the  resistance  has 
risen  to  6.32  ohms.  What  is  temperature  rise  of  the  coil? 

The  equation 

Rt  =  R25(l  +  .  00381 T) 

can  be  transformed  to 

T->    r> 

rn  _ 


.00381#25 ' 
when     Rt  =  6.32  =  resistance  at  high  temperature ; 

D        K     AQ  U  .  i       OPio    O    • 

Hi  25 —  9,'xa —  ^O      \J.  , 


6.32-5.48 

~ 


.  00381X5.48' 


Temperature  rise  therefore  is  40.3°  C. 

Problem  22-5.  The  cold  (20°  C.)  resistance  of  an  armature  was 
2.18  ohms.  The  hot  resistance  was  2.56  ohms.  What  was  the 
temperature  rise? 

Problem  23-5.  An  electric  soldering  iron  has  a  resistance  com- 
posed of  iron  wire  of  80  ohms  when  cold  (20°  C.).  When  hot 
the  resistance  rises  to  150  ohms.  What  temperature  is  reached, 
assuming  a  constant  temperature  coefficient? 


104  ELEMENTS  OF  ELECTRICITY 

78.  Temperature  Coefficient  of  Alloys,  etc.      It  has  been 
stated  that  the  temperature  coefficient  of  resistance  for  all 
pure  metals  is  nearly  the  same,  that  is,  somewhere  about 
.004.     Alloys,  though  of  a  much  higher  resistance  per  mil-ft. 
have  much  lower  coefficients,  some  having  "  O  "  and  even 
negative  coefficients  at  certain  temperatures. 

"  Manganin,"  an  alloy  consisting  of  copper,  nickel,  and 
iron-manganese,  for  instance,  has  a  resistance  per  mil-ft. 
of  from  250  to  450  ohms  according  to  the  proportions  of 
the  different  metals  used,  and  a  temperature  coefficient  so 
low  as  to  be  practically  negligible. 

Certain  substances,  notably  carbon,  porcelain,  glass,  etc., 
decrease  in  resistance  very  rapidly  when  heated.  The 
cold  resistance  of  a  carbon  lamp  filament  is  about  twice 
as  great  as  the  "  hot  "  resistance.  The  porcelain  ' i  glower  " 
of  a  Nernst  lamp  when  cold  is  a  good  insulator,  but  when 
heated  to  incandescence  it  becomes  a  conductor.  The 
filaments  of  the  new  tungsten  lamps  are  pure  metal  and 
accordingly  have  a  positive  coefficient,  which  is  about  .0051. 

79.  Copper- Wire   Tables.       Tables  have  been  prepared 
by  wire  manufacturers  and  by  the  A.I.E.E.  which  giVe  the 
resistance  of  1000  ft.  or  of  a  mile  of  copper  wire  of  different 
standard  sizes,   and  several  temperatures.     The  sizes  are 
designated  by  gauge  numbers,  diameter  in  mils,  and  sec- 
tion area  in  cir.  mils,  etc.     There  are  several  standard  wire 
gauges,  B.  &  S.  (Brown  &  Sharpe)  in  general  use  in  America, 
B.  W.  G.  (Birmingham  Wire  Gauge)  in  general  use  in  Great 
Britain.     See  Appendix  for  B.  &  S.  table. 

By  means  of  these  tables  it  is  very  easy  to  find  the  re- 
sistance of  any  length  of  wire  of  a  given  section  area,  etc. 

Example.     What  copper  wire  (B.  &  S.  gauge)  should  be  used  to 
transmit  electric  power  2  miles  (out  and  back) ;  resistance  not  to 
exceed  2.7  ohms;  temperature  to  be  assumed,  20°  C.? 
2  miles  =  2X5280  =  10,560  ft. 

2.7 

2.7  ohms  for  2  miles  =  r^~  ohms  per  thousand  ft. 
lU.oo 

=   .256  ohms  per  thousand  ft. 


MEASUREMENT  OP  RESISTANCE  105 

From  wire  table  : 

No.  5  =  .3128  ohm  per  thousand  ft.; 
No.  4  =  .2480    "      " 

No.  4  must  be  used  in  order  not  to  exceed  limit  of  .256  ohm 
per  thousand  ft. 

Example.  Suppose  line  in  above  Example  is  to  work  at  a 
temperature  of  45°  C.,  what  number  wire  will  be  required? 

The  resistance  is  not  given  in  the  tables  at  45°  C.,  but  at  20°  C. 
We  must  then  find  out  what  resistance  a  wire  will  have  at  20°  C., 
which  has  2.7  ohms  at  45°  C. 

According  to  the  equation, 


From  above  table: 

a  (for  45°)  =  .00353: 
T  =  45°-20°  =  25°; 
#20  =  2.7  (1-.  00353X25); 
=  2.7(1  -.0883) 

=  2.46  ohms  for  the  two  miles; 
or 

.233  ohm  per  thousand  ft. 

The  problem  then  becomes:   What  size  wire  has  .233  ohm  per 
thousand  ft.  at  20°  C.? 
From  table  : 

No.  4  has  .248  ohm  per  thousand  ft. 
No.  3    "    .1967    "      " 

No.  3  must  be  used  in  order  not  to  exceed  limit  of  .233  ohm 
per  thousand  ft. 

Problem  24-6.  What  size  wire  (B.  &  S.)  is  that  mentioned  in 
Problem  8-5? 

Problem  25-6.  What  size  is  trolley  wire  in  Problem  11-5? 

Problem  26-5.  What  size  wire  must  be  used  to  transmit  40 
amperes  from  generator  to  lamps,  a  distance  of  600  ft.  with  2  volts 
drop  in  line?  Line  to  work  at  30°  C. 

80.  Standard    Ohm.     Measurement  of    Resistance.     The 

standard  ohm  has  already  been  defined  as    the   resistance 


106  ELEMENTS  OF  ELECTRICITY 

of  a  column  of  pure  mercury  at  0°  C.,  106.3  cms.  long,  of 
uniform  cross-section  and  weighing  14.4521  gms. 

This  piece  of  apparatus  is  difficult  to  construct  and  to 
operate,  so  several  secondary  standards  have  been  adopted, 
such  as  a  copper  coil  in  oil.  The  value  of  an  unknown  re- 
sistance can  be  found  by  comparing  it  with  one  of  these 
secondary  standards,  by  means  of  the  following  methods: 

FALL  OF  POTENTIAL 

There  are  needed  for  this  method :  (1)  a  known  or  standard 
resistance  R,  (2)  a  voltmeter  (V),  and  (3)  a  source  of  power. 
See  Fig.  104. 

The  known  resistance  (R)  and  unknown  (Rx)  are  con- 
nected to  the  power  in  series,  so  that  there  will  be  the  same 

current  flowing  through  each. 
The  voltmeter  is  placed  to  read 
the  voltage  (vi)  across  R,  and 
then  to  read  the  voltage  (v2) 
across  Rx.  Since  the  same  cur- 
rent is  flowing  in  each  resist- 

FIG.  104.  Method  of  measuring  re- 

sistance  by  voltmeter  and  stand-    anC6,    the     Voltage     acrOSS     each 
ard  resistance.  .         .  .  . 

must  be  in   the   same   ratio  as 

the  two  resistances.  That  is,  if  the  voltage  v2  is  twice 
the  voltage  v];  then  the  resistance  Rx  must  be  twice 
R,  for  we  have  seen  that  it  requires  twice  the  voltage  to 
force  the  same  current  through  twice  the  resistance.  We 
place  the  coils  in  series  in  order  to  be  sure  that  there  is  the 
same  current  in  each.  We  have  seen  in  Chapter  III  that 
this  relation  may  be  expressed  as  an  equation : 


R 
Thus, 


VR 


MEASUREMENT   OF  RESISTANCE  107 

Since  R,  t>2,  and  i'i  are  known  quantities,  the  value  of  the 
unknown  resistance  (Rx)  may  be  found. 
The  advantages  of  this  method  are : 

(1)  The   voltmeter   need   not   be   accurately   calibrated, 
providing  the  deflections  are  proportional  to  the  voltage. 

(2)  It  is  an  accurate  method  of  measuring  low  resistances 
if  the  voltmeter  has  a  high  resistance,  or  is  replaced  by  'a 
galvanometer.     This  shunts  very  little  current  around  the 
series  resistance. 

(3)  By  means  of  a  potentiometer  in  place  of  the  volt- 
metert  no  current  is  shunted  around  the  resistance  and  thus 
very  low  resistances  can  be  measured  accurately. 

'  81.  Voltmeter — Ammeter.  An  unknown  resistance  (R)  may 
be  placed  in  series  with  an  ammeter  (A)  and  the  voltage  across 

it  be  found  by  means  of  a  voltmeter  (F),  Fig.  105.     Then  R  =  J 

(Ohm's  law).  This  requires  that  both  ammeter  and  voltmeter  be 
accurately  calibrated.  The  voltmeter  must  also  be  of  high  re- 
sistance. If  there  is  not  much 
current  in  the  circuit,  the  volt- 
meter is  generally  connected  around 
both  R  and  A.  The  ammeter 
would  then  be  a,  millammeter .  If 
a  millammeter  were  connected  as 
in  diagram,  it  would  read  the 
sum  of  the  currents  going  through  FlG  105.-Method  of  measuring  re- 

both  R  and  F,  which  WOUld  be   ap-        sistance  by  voltmeter  and  ammeter. 

preciably  larger  than  that  through 

R  alone.  On  the  other  hand,  if  the  voltmeter  were  put  across  R 
and  A,  the  fall  across  A  would  be  too  small  to  be  indicated  and 
thus  only  the  voltage  across  R  would  be  read.  See  Chapter  XIV. 

This  method  has  the  advantage  of  making  use  of  the  com- 
mon instruments,  ammeter  and  voltmeter,  and  is  used 
wherever  extreme  accuracy  is  not  desired. 

82.  Wheatstone  Bridge.  By  far  the  most  important 
apparatus  for  accurately  measuring  resistance  is  the  well- 
known  Wheatstone  bridge. 

It  consists  fundamentally  of  a  loop  of  four  resistances, 
R,  RI,  R2,  RZ,  one  of  which  is  unknown,  Fig.  106.  An 


108  ELEMENTS  OF   ELECTRICITY 

electric  current,  generally  from  a  battery,  is  sent  into  the 
loop  at  B,  and  immediately  divides  into  two  parts,  one  part 
taking  the  branch  composed  of  the  resistances  R  and  R2, 
the  other  part  taking  the  branch  made  up  of  RI  and 
R$,  Both  branches  come  together  again  at  D,  and  return 
the  current  to  the  battery.  A  galvanometer  G  is  now 
"  bridged  "  between  the  two  branches  from  A  to  C. 

Assume  R%  to  be  the  unknown  resistance.  R,  RI,  R2 
are  variable  resistances,  and  are  now  adjusted  until,  on 
closing  the  key  in  the  circuit  of  the  galvanometer,  it  gives 

no  deflection.  This  is  called 
"  balancing "  the  bridge. 
The  points  A  .and  C  are  then 
at  the*  same  level  or  poten- 
tial. For  we  have  seen  that 
where  there  is  any  difference 
D  of  potential  between  two 

FIG.  106.-Wheatst<me  bridge,  diamond      pointS)       a      current      always 

flows  when   they  are  joined 

by  a  conductor.  Since  the  points  A  and  C  are  joined 
through  the-  galvanometer,  and  no  current  flows,  they 
must  be  at  the  same  potential. 

Let  us  now  consider  the  voltages  across  the  separate 
resistances.  We  have  said  that  the  current  divides  into 
two  branches,  BAD  and  BCD.  Now  the  current  in  each 
branch  need  not  be  the  same,  so  let  us  designate  the  cur- 
rent in  the  branch  BAD,  by  the  letter  /,  and  the  current 
in  the  branch  BCD,  by  the  letter  /i. 

Then  the  voltage  across  BA  (that  is,  R)  =IR; 
"       "         "  "      BC  (that  is,  RI)  =/i#i. 

But  we  have  said  that  the  points  A  and  C  are  at  the  same 
level  (since  no  current  flows  through  the  galvanometer) 
and  therefore  the  voltage  drop  from  B  to  A  (IR)  =the 
voltage  drop  from  B  to  C  (/i#i).  That  is, 


MEASUREMENT  OF  RESISTANCE  109 

For  the  same  reasons,  the  voltage  drop  from  A  to  D, 
(7/?2)  must  equal  the  voltage  drop  from  C  to  D  (I\R^)  ; 
that  is, 

(2)  IR2=IiR9. 

Dividing  equation  (1)  by  equation  (2)  we  get, 


Cancelling  the  /'s  and  /i's, 


This  is  the  fundamental  equation  of  the  Wheatstone  bridge. 
Since  R^  is  the  only  unknown,  the  equation  can  be  solved 
and  the  value  of  R%  found. 

~D 

The  ratio  -~-  is  generally  made  to  equal  some  such  fraction 
#2 

7?        1 

as  -j1^,  and  then  it  follows  that  -^=77:.      R\  is  called  the 

-n/3      1U 

Rheostat  arm.  This  makes  #3  10  times  the  value  of  R\.  If 
R2  is  made  the  same  as  R}  then  R%  will  be  the  same  as  RI. 

In  -writing  the  equation  for  the  bridge,  notice  that  we 
started  at  B  (a  point  where  the  battery  line  came  in). 
For  one  side  of  our  equation  we  read  along  one  bianch. 

T) 

That  is,  we  said,    "R  is  to  R2  "  or  -=-.     For  the  other  side 

/12 

of  the  equation  we  went  back  to  the  point  B  again  and  read 

D 

along  the  other  branch,  saying,    "Ri  is  to  #3  "  or  -^.     We 

did  not  continue  around  the  loop  and  make  the  second  side 
of  the  equation  "R^  is  to  RI."  Had  we  done  so,  the  two 
ratios  would  have  been  unequal  and  we  could  have  formed 
no  equation. 


110  ELEMENTS  OF  ELECTRICITY 

It  is  to  be  noticed  then,  that  in  making  the  fundamental 
equation  for  a  Wheatstone  bridge,  we  read  along  one  branch 
for  one  side  of  the  equation,  then  go  back  and  read  along  the 
other  branch  for  the  other  side  of  the  equation. 

It  is  interesting,  and  of  value,  to  know  that  it  makes  no  difference 
at  what  point  we  start,  providing  we  start  at  the  same  point  for 
each  side  of  the  equation.  Thus  suppose  we  start  at  A.  Reading 

7-> 

along  the  upper  branch,  we  have,   "R  is  to  R ^  "  or—,  for  one  side 

Hl 

of  the  equation.  Then  coming  back  to  the  point  A  and  reading 
the  lower  branch  we  have,  "R2  is  to  R3."  The  equation,  then, 
/?  7? 

—  =  Tr  is  Just  as  true  as  the  first  and  as  easy  to  solve. 
RI     RS 

Tf 

If  we  start  at  the  point  C,  we  have   "R3  is  to  R2  "  or  — -,  "#! 

R2 

7?  7?        7? 

isto/2,"  or  -=— *,  that  is,  ^~  =  -~,  which  equation  is  also  true. 
R  HI     R 

The  important  details  about  measuring  resistance  with  a  Wheat- 
stone  bridge  are : 

(1)  Make  a  complete  loop  of  four  resistances. 

(2)  Connect   battery   terminals    (through  key)   to  two  points 
that  are  not  adjacent,  that  is,  each  branch  of  the  current  must 
always  flow  through  two  resistances  before  returning  to  the  battery. 

(3)  Connect  galvanometer  through  key  across  the  two  remain- 
ing points. 

(4)  In  forming  the  equation,  start  at  any  point  and  read  along 
two  resistances  for  one  side  of  the  equation;   then  come  back  to 
same  point,  and  read  along  the  other  two  resistances  for  the  other 
side  of  the  equation. 

Problem  27-6.  In  a  Wheatstone  bridge  arranged  as  in  Fig. 
106,  a  balance  is  obtained  when  R  =  W  ohms,  7^=45  ohms,  R2  = 
100.  What  is  the  value  of  R3? 

r> 

Problem  28-5.     In  Fig.  106,  if  —  =  100,  and  R3  =  4.2Q,  what 

R2 

is  the  value  of  RJ 

Problem  29-5.  What  value  must  R  2  have,  if  R  is  1000  ohms 
and  R!  is  48  ohms  and  it  is  desired  to  measure  R3,  which  is  known 
to  be  about  500  ohms? 

83.  Instructions  for  Use  of  Wheatstone  Bridge. 

1 .  Never  hold  the  bridge  plugs  in  the  hand  or  place  them  where 
they  are  liable  to  come  in  contact  with  oils  or  acids.  They  may 


MEASUREMENT  OF  RESISTANCE  111 

become  covered  with  oil  and  the   "plug  resistance  "  thus  increased, 
or  may  be  corroded  so  that  they  do  not  properly  fit  the  sockets. 

2.  Insert  all  plugs  firmly,   with  a  twisting  motion.     Do  not, 
however,  insert  them  with  force  enough  to  strain  the  top. 

3.  Always  put  a  key  in  the  battery  circuit  and  one  in  the  gal- 
vanometer circuit.     Close  the  battery  circuit  first,  then  the  circuit 
through  the  gvalanometer.     In  breaking  the  circuit,  open  the  key 
in  the  galvanometer  circuit  first. 

4.  In  closing  the  galvanometer  circuit,  make  only  slight  contact 
with  the  key  until  balance  is  nearly  secured.     The  galvanometer 
may  thus  be  protected  from  heavy  currents.     If  the  galvanometer 
is  very  sensitive,  put  a  shunt  across  its  terminals  until  the  bridge 
is  nearly  balanced. 

5.  In  making  a  resistance  measurement  proceed  in  the  same 
manner  as  in  weighing  with  a  chemical  balance.    First  put  in  a  coil 
estimated  to  be  about  correct  for  the  resistance  being  measured. 
If  this  is  too  low,  use  one  twice  as  great,  if  too  high,  one  one-half 
as  great.     Fix  in  this  way  two  limits  between  which  the  resistance 
lies.     Then  systematically  bring  these  limiting  values  closer  and 
closer  together,  until  the  nearest  balance  on  the  bridge  is  obtained. 
It  will  not,  in  general,  be  possible  to  secure  an  exact  balance,  but 
two  values  for  the  resistance  in  the  rheostat  arm  may  be  found 
which  will  give  steady  deflections  of  the  galvanometer  needle 
in  opposite  directions.     The  correct  value  lies  between.     From 
the  deflections  of  the  galvanometer  needle  in  the  two  cases,  inter- 
polate for  the  resistance  which  will  give  no  deflection.     Thus,  sup- 
pose the  smallest  coil  on  the  bridge  is  0.1  ohm.     With  this  added, 
a  steady  deflection  of  the  galvanometer  of  2  divisions  to  the  right 
is  given.     Without  it,  the  deflection  is  3  divisions  to  the  left. 
0.1  ohm  makes  a  difference,  therefore,  of  5  divisions,  and  the  re- 
sistance which  would  give  no  deflection  is  therefore  1X0.1=0.06 
ohm. 

6.  When  balance  has  been  obtained,  reverse  the  current  through 
the  bridge  and  see  if  the  balance  is  maintained. 


84.  Slide-wire  Bridge.  Sometimes  a  single  straight 
wire  of  uniform  section  and  of  high  resistance  takes  the 
place  of  the  resistances  R  and  ^2  as  in  Fig.  107.  The  point 
A  is  then  a  sliding  contact  piece.  R  is  the  resistance  of  the 
straight  wire  from  B  to  A,  and  R%  the  resistance  from  A 
to  D.  In  this  form  the  bridge  is  called  the  slide-wire  bridge; 
a  balance  being  obtained  by  sliding  the  contact  maker 
A  along  the  wire  BD,  thus  decreasing  or  increasing  R  or 
#2  as  desired.  The  ratios  are  read  as  before,  the  lengths 


112 


ELEMENTS  OF   ELECTRICITY 


BA  and  AD  being  used  instead  of  the  actual  resistances, 

since  the  resistance  is  proportional  to  the  length. 

This  form  has  the  advantage  that  a  balance  can  be  obtained 

very  quickly.     It  is  not  as  accurate  as  the  other  forms, 

however,  because  of  the  wear- 
ing of  the  wire,  and  of  the 
necessarily  low  resistance  of 
R  and  R%. 

Sometimes  R\  is  a  perma- 
nent resistance  coil,  and  the 
wire  is  so  calibrated  that  the 
resistance  of  R3  is  read  di- 
rectly from  the  position  of 
the  slider  A  on  the  wire.  It 
is  then  called  an  "  Ohm- 
meter."  A  telephone  re- 
ceiver in  this  case  generally 

takes  the  place  of  the  galvanometer. 

Resistance  coils  are  often  placed  in  R  and  R2  to  increase 

the  accuracy  of  this  form  of  bridge,  though  by  so  doing  the 

range  is  lessened. 

Problem  30-5.  In  a  bridge  arranged  as  in  Fig.  107,  BD  is  100 
cms.  long,  R!  5  ohms.  If  bridge  balances  when  BA  is  37.5  cms., 
what  value  is  -R3? 


FIG.  107. — Wheatstone  bridge,  slide- 
wire  form. 


Problem  31-5.     If  7^  =  16  and 

the  values  of  B  A  and  AD  be?     BD 


3  =  90  (Fig.  107),  what  must 
100  cms. 


85.  Location  of  Faults  in  Cables,  etc.  It  often  happens  that 
a  telephone  conductor  becomes  "  grounded."  It  is  then  very  de- 
sirable to  locate  the  distance  of  this  "  fault  "  from  one  end  or  the 
other  of  the  line  before  starting  out  from  the  office  to  repair  it. 
Where  the  wire  is  laid  underground,  it  would  be  too  costly  a  pro- 
ceeding to  inspect  any  great  length  of  cable,  and  thus  very  refined 
methods  have  been  worked  out  for  locating  such  faults  to  a  great 
degree  of  precision.  There  are  two  methods  in  common  use, 
called  the  Murray  loop  and  the  Varley  loop.  They  are  both  modi- 
fications of  the  Wheatstone  bridge,  and  therefore  the  underlying 
principles  stated  above  apply  to  them, 


MEASUREMENT   OF  RESISTANCE 


113 


86.  Murray  Loop.  This  test  ordinarily  makes  use  of  the  slide- 
wire  form  of  the  bridge  and  is  arranged  as  in  Fig.  108. 

There  is  a  "  ground  "  at  F  in  the  conductor  DC,  of  a  line  between 
two  cities.  A  good  wire  BC,  which  happens  to  lie  between  the 
two  cities  and  to  be  exactly  like  DC,  is  joined  to  DC  at  C.  By 
means  of  the  slide-wire  BD,  a  complete  loop  of  resistances  is 
formed.  The  battery  is  connected  to  the  slider  A  and  to  the 
ground.  This  is  the  same  as  connecting  in  the  battery  at  A  and 
F,  since  the  ground  serves  as  a  conductor.  The  galvanometer  is 
placed  across  BD.  We  now  have  a  regular  slide-wire  Wheatstone 
bridge,  as  in  Fig.  107;  the  resistance  of  the  slide-wire  R2  and  R 
forming  one  side  of  the  bridge,  while  R3  (the  resistance  out  to 
the  fault)  and  RI  (the  resistance  through  good  wire  to  fault)  form 
the  other  side. 


FIG.  108. 


The  bridge  is  balanced  by  moving  A  along  the  wire  till  there 
is  no  deflection  of  the  galvanometer.  Then  the  regular  equation 
of  the  Wheatstone  bridge  holds  true: 


R  ~~R.L' 

If  we  let  L  equal  the  length  of  entire  loop,  which  would  be  known, 
and  let  x  equal  distance  out  to  fault,  then  the  equation  would 
become, 


=__ 
R     L-x 

As  x  is  the  only  unknown  quantity,  it  can  easily  be  found  as  fol- 
lows: 


But 


=  the  entire  length  of  the  slide-wire, 


114 


ELEMENTS  OF  ELECTRICITY 


This  shows  that  x  is  the  same  fraction  of  the  total  length  of  the 
loop  that  R2  is  of  the  total  length  of  the  slide-wire. 

If  the  "  good  "  wire  BC  is  not  of  the  same  size  and  material  as 
the  "faulty"  wire,  the  resistance  of  the  loop  BCD  is  first  found 
in  the  usual  way  by  a  bridge.  It  is  then  connected  up  as  in  Fig. 
108,  and  worked  out  as  above,  with  the  exception  that  L  stands 
for  the  resistance  of  the  entire  loop  instead  of  for  the  length  of  it, 
and  x  for  the  resistance  out  to  fault.  If  the  size  of  the  "  faulty  " 
wire  is  known,  the  distance  out  to  the  fault  is  readily  computed. 

In  the  place  of  the  slide-wire  BD,  two  sets  of  resistance  coils 
are  often  substituted. 

87.  Varley  Loop.  In  the  Varley  loop,  Fig.  109,  instead  of  vary- 
ing the  ratio  of  R2  to  R  as  in  the  Murray  form  of  bridge,  these 


FIG.  109.  —  Varley  loop  for  locating  fault. 

resistances  are  left  constant  and  a  balance  is  obtained  by  means 
of  a  third  set  of  resistance  coils  R3  inserted  in  series  with  the 
faulted  cable.  The  resistance  of  the  total  length  of  the  good  and 
bad  cables  L,  is  either  known  or  first  found  as  in  Murray  loop. 
The  value  of  x  (resistance  out  to  fault)  is  then  found  by  the 
equation  of  the  Wheatstone  bridge  : 


R~  L-x' 
R2L-RR3 


As  all  the  values  in  the  right-hand  member  of  the  equation  are 
known,  the  value  of  x  may  be  found  and  the  distance  out  to  the 
fault  F  be  computed  from  the  resistance  per  foot  of  the  cable. 

An  interesting  variation  in  the  use  of  the  Varley  loop  is  shown 
\n  Fig.  110. 


MEASUREMENT   OF  RESISTANCE 


115 


The  two  cables  must  here  be  "  twin  "  wires.  Then  N  repre- 
sents the  resistance  out  to  fault  on  the  bad  wire.  Thus  N  would 
also  represent  the  resistance  of  an  equal  distance  out  on  the  good 
cable.  Now  let  x  represent  the  resistance  of  the  bad  cable  from 
fault  F  to  far  end.  Then  x  will  also  represent  the  resistance  of 
an  equal  length  of  the  good  cable.  The  resistances  72  2  and  R  are 
made  equal  and  the  bridge  is  balanced  by  regulating  R3  as  before. 
The  equation  then  becomes: 


R 


But  as  R  2  =  R,  we  have, 


FIG.  110. — Varley  loop;  for  locating  a  fault  in  "twin  cables." 

Thus  x  equals  £  of  whatever  we  have  to  make  R3  in  order  to 
balance  the  bridge.  This  gives  a  very  rapid  method  of  finding 
the  location  of  a  ground  in  a  telephone  circuit,  and  is  as  accurate 
as  the  resistances  used  in  R}  R  2  and  R «. 

Problem  32-5.  A  Murray  loop  is  arranged  as  in  Fig.  108  with 
a  meter  slide-wire.  BA=Q5  cms.  when  bridge  is  balanced.  The 
length  of  the  line  between  the  two  cities  is  2  miles.  The  return 
wire  is  exactly  like  the  faulted  wire.  How  far  from  test  end  is 
fault  F? 

Problem  33-5.  In  a  Murray  loop  arranged  as  in  Fig.  108, 
the  resistance  of  the  loop  =  44. 2  ohms.  The  line  wire  DC  is  No. 
18  B.  &  S.  copper.  If  bridge  balances  with  AD  equal  to  32  cms., 
how  far  from  test  end  is  fault  Ft 

Problem  34-5.  A  Varley  loop  in  which  the  resistance  of  the 
twin  cables  is  44.8  ohms  is  arranged  as  in  Fig.  109.  A  balance  is 


116  ELEMENTS  OF   ELECTRICITY 

obtained  when  R=  10  ohms,  R2  — 100  ohms,  7?3  =  357  ohms.     If 
the  faulted  wire  is  copper,  No.  14  (B.  &  S.)  how  far  from  E  is  fault? 

Problem  35-5.  A  Varley  loop  is  arranged  as  in  Fig.  110  with 
the  return  the  same  as  the  faulted  line,  No.  12  (B.  &  S.)  copper. 
R  =  RZ;R3  =  4.12  ohms.  How  far  from  C  is  fault  F? 

Problem  36-5.  If  wire  in  Problem  35-5  is  iron  and  50  ohms  to 
the  mile,  howr  far  will  F  be  from  C? 

88.  Voltmeter  Method  for  High  Resistances.  It  is  often 
desirable  to  know  whether  a  certain  resistance  comes  within 
or  exceeds  a  given  value;  as,  for  instance,  the  insulation 
resistance  between  the  commutator  and  the  frame  of  a 
motor.  A  voltmeter  of  known,  resistance  Rv  is  placed 

across  a  constant  source  of 
power  as  A  B,  Fig.  Ill,  and 
voltmeter  reading  v  is  noted. 

The  frame  of  the  motor  MF 
is  then  connected  to  terminal 
A,  and  commutator  segment 
C  is  connected  through  the 

FIG.     111.— Voltmeter    method    -of     ^U™^™     f0     fV,A     n+Vi^r-    tm-mi 
measuring  insulation  resistance.        Voltmeter 

nal  B.     The  reading    Vi    of   the 

voltmeter  is  then  taken.  Since  the  motor-insulation  and 
voltmeter  are  in  series,  the  voltage  of  A  B  is  distributed 
across  the  motor  and  voltmeter  in  direct  proportion  to  their 
separate  resistances.  The  voltmeter  reads  the  voltage 
across  itself  as  always,  thus  v{  =,voltage  across  the  voltmeter. 
The  voltage  across  the  motor-insulation  then  equals  the 
voltage  across  AB,  v,  minus  voltage  across  voltmeter,  v\. 
That  is, 

Voltage  across  voltmeter  =vjj 
Voltage  across  motor-insulation  =v  —  vi. 

Since  the  two  are  in  series,  the  resistances  of  each  are  propor- 
tional to  the  voltages  across  each. 

Insulation  R  of  motor  _v  —  v\  (volts  across  motor-insulation) 
R  of  voltmeter  v\  (volts  across  voltmeter) 


MEASUREMENT   OF  RESISTANCE  117 

Let  x  =  (insulation)    resistance  of  motor.    Equation  then 
becomes, 

x      v  —  v± 


This  equation  can  be  solved  for  x,  since  all  other  values 
are  known. 


A  deflection  of  1  volt  for  vi  would  give  x  the  following 
value  : 

x=Rv(v-l). 

If  no  deflection  is  noticed  when  the  voltmeter  is  con- 
nected in  series  with  motor-insulation,  we  know  that  the 
insulation  is  at  least  greater  than  Rv(v—l).  Thus  if  the 
resistance  of  the  voltmeter  Rv  is  15,000  ohms,  and  the  volt- 
age across  AB,  (v),  is  111  volts,  we  would  know,  that  in  case 
of  no  deflection,  we  had  at  least  15,000X110=1,650,000 
ohms  insulation  resistance  between  frame  and  commutator. 

89.  Insulation  Resistance  of  Covered  Wire,  etc.      It  is  re- 

quired by  fire  insurance  companies  that  insulated  wire  of  various 

grades  shall  be  up  to  a  certain  standard  insulation  resistance  per  mile 

after  being  soaked  in  salt  water 

for  a  stated  length  of  time.     The  10o,oooohm8 

following  method  is  the  common 

practice    for    finding   the    ohmic 

resistance  of  the  insulation.     In 

Fig.  112  a  known  high  resistance,        FlG  112._insuiation  resistance. 

as    100,000  Ohms,  IS  Connected  in  Standardizing  the  galvanometer. 

series  with  a  shunted  galvanom- 

eter G  across  a  constant  potential  source  of  power  AB.  The 
resistance  of  the  galvanometer  and  of  the  shunt  S  are  known. 
The  deflection  of  the  shunted  galvanometer  is  noted,  and  by 
means  of  the  resistances  of  the  galvanometer  and  shunt  it  is  possi- 
ble to  compute  what  the  deflection  would  be  without  the  shunt. 
For  example,  suppose  the  following  data  to  be  obtained  : 

Deflection  =     80  scale  divisions. 

Resistance  of  (r  =  2000  ohms. 

"  S  =     10      " 


118  ELEMENTS   OF    ELECTRICITY- 

Since  the  S  and  G  are  in  parallel,  the  current  will  divide  between 
in  inverse  proportion  to  the  resistance;  that  is,  the  greater  part 
of  the  current  will  go  through  the  smaller  resistance  S. 

Assume  the  current  to  be  divided  up  into  2010  parts;  then 
2000  parts  flow  through  the  shunt  and  10  parts  through  G.  That 
is,  Y£TO  parts  of  the  current  go  through  the  galvanometer,  and 
therefore  the  deflection  of  80  is  only  2^fo  (or  T£T),  as  large  as  it 
would  be  if  it  were  unshunted. 

An  unshunted  galvanometer  would  then  deflect  201  X  80  =  16,080 
divisions,  when  connected  in  series  with  100,000  ohms,  and  the 
same  current  were  flowing  in  the  line.  If  there  were  1,000,000 
ohms  (1  megohm)  in  series,  the  unshunted  galvanometer  would 
deflect  only  TO  as  much  (the  more  resistance  the  less  current). 

That  is,  the  deflection  would  be  —  X 16080  =  1608  (the  deflection 

for  1  megohm).     This  is  called  the   galvanometer   constant.     The 

shunt  is  now  removed  from  the  gal- 
vanometer as  in  Fig.  113.  The  coil 

07  LGJ        ^  immersed  in  salt  water,  is  put  in 

*      ~a  place  of  the   100,000  ohms,  one   con- 

nection being  made  to  the  end  y  of 
the  coil,  and  the  other  to  the  metal 
tub.  The  other  end  x  of  the  coil  is 
kept  above  the  water,  so  that  any 

FIG.  iis.-lnsulation  resis-       current  which  goes  through  G  must 
tance.  leak   through   insulation   of   the  wire 

coil  into  the  water. 

Assume  the  galvanometer  now  reads  16.  Since  a  deflection 
of  1608  means  a  megohm  resistance  in  the  line,  a  deflection  of 
only  16  means  a  much  greater  than  1  megohm  is  in  the  line.  In 
fact,  the  deflections  are  inversely  proportional  to  the  resistance 
of  the  line.  Thus  the  unknown  insulation  resistance  x  is  as  much 
greater  than  one  megohm  as  1608  is  greater  than  16. 

x  = =  100.5  megohms. 

lo 

This  is  the  insulation  resistance  of  the  coil. 

If  the  coil  were  \  mile  long  the  insulation  resistance  per  mile 
would  be  only  \  as  great,  or  25.1  megohms.  This  is  because  the 
current  would  have  4  times  as  much  area  to  leak  through  in  a 
mile  as  in  \  mile. 

Notice  that  the  galvanometer  constant  1608  is  not  used  as  a 
multiplier  but  as  a  dividend,  which  is  divided  by  the  deflections. 

Note  that  in  this  test  the  resistance  of  the  battery  and  gal- 
vanometer are  neglected  because  they  are  too  small  in  comparison 
with  the  resistance  of  the  insulation  and  standard  box  to  appreciably 
affect  the  results.  When  the  standard  100,000-ohm  box  above 


MEASUREMENT   OF   RESISTANCE  119 

is  in  the  line,  the  galvanometer  is  shunted  so  that  it  has  less  than 
10  ohms  resistance,  which  is  one  ten-thousandth  of  the  total  re- 
sistance of  the  circuit.  When  unshunted  in  the  second  part  of  the 
test,  the  resistance  of  the  galvanometer  is  but  one  fifty-thousandth 
of  the  total  line  resistance.  In  fact,  the  standard  resistance  of 
100,000  ohms  might  be  left  in  the  line  in  the  second  part  as  well, 
and  its  resistance  neglected,  so  small  is  it  in  comparison  with  the 
insulation  resistance. 

Problem  37-5.     In  a  test,  as  in  Paragraph  89,  for  the  insula- 
tion of  a  coil  of  wire  the  following  data  were  taken: 
Resistance  of     Gal.  =  3000  ohms ; 
"  "   Shunt  =  200  ohms. 

Deflection  of  shunted  Gal.  with  200,000  ohms  in  series  =  18.4. 
Deflection  of  unshunted  Gal.  with  the  resistance  of  the  insula- 
tion in  series  =  4. 3. 

Length  of  coil  =  500  ft. 

(a)  What  is  the  galvanometer   "  constant?  " 

tf     tt    u    insuiation  resistance  per  mile  of  cable? 

90.  Construction    of   Various    Types    of   Bridges.      The 

appearance  of  the  regular  "  Post-office "  form  of  Wheat- 
stone  bridge  is  shown  in  Fig.  114. 


FIG.  114. — Wheatstone  bridge,  "post-office"  form. 

The  resistance  wire  is  wound  non-inductively  on  spools 
as  in  Fig.  116.  The  ends  of  the  wire  are  now  connected 
across  the  gap  between  two  adjacent  brass  blocks,  as  shown 


120 


ELEMENTS  OF  ELECTRICITY 


in  Figs.  115  and  116.     The  construction  of  one  such  coil  and 


LINE    <$) 


FIG.  115. — Diagram  of  connections  of  P.  O.  form  of  Wheatstone  bridge. 

set  of  blocks  is  shown  in  Fig.  116.     When  the  brass  plug 

is  inserted  in  the  tapered  hole 
between  the  blocks,  it  forms 
a  short  circuit  around  the 
coil  and  "  cuts  the  coil  out  " 
of  the  circuit.  In  order  to 
put  a  certain  amount  of  re- 
sistance into  a  circuit,  say 
between  z  and  y,  Fig.  115, 
it  is  merely  necessary  to  re- 
move the  plug,  which  short- 
circuits  a  coil  of  the  desired 
resistance.  Thus  the  plug 
being  removed  from  hole  C, 
the  current  must  go  through 
the  coil  C  of  10  ohms  in 
flowing  from  z  to  y  or  vice 
versa.  Binding  posts  are 
connected  to  certain  of  the 
blocks  as  x,  y,  z  and  t, 

FIG.  116. — Details  of  connection  of  non-     SO  that  the    bridge    Can    be  di- 
inductive  coil  to  blocks  on  bridge  top. 

vided  into  three  parts.     Thus 
from  z  to  y  may  be  one  ratio,  from  x  to  z  the  other  ratio. 


MEASUREMENT  OF  RESISTANCE 


121 


and  from  x  to  t,  the  variable  resistance  by  which  the 
bridge  is  balanced.  Of  course  the  unknown  resistance  has 
to  be  connected  across  the  end  from  y  to  t  of  the  bridge  in 
order  to  complete  the  loop  of  four  resistances. 

In  Fig.  117  is  shown  the  appearance  of  the  "  Decade  " 
form  of  bridge.     In  this  type,  resistance  is  introduced  into 


.FiG.  117. — Wheatstone  bridge,  "decade"  form. 

a  circuit  by  placing  the  plug  in  the  hole,  instead  of  by  re- 
moving the  plug.  The  details  of  the  construction  are 
shown  in  Fig.  118. 

The  coils  composing  the  variable  resistance  for  balancing 
the  bridge  are  arranged  in  such  an  ingenious  way  that  10 
variations  in  value  in  each  "  decade  "  or  row  are  made  by 
the  use  of  only  4  coils.  The  diagram  shows  the  details 


122 


ELEMENTS  OF  ELECTRICITY 


of  this  scheme.  The  student  will  be  interested  in  studying 
out  for  himself  how  the  different  values  are  obtained  by 
"  plugging  in  "  at  the  proper  place.  In  a  bridge  where 
there  are  four  "  decades/'  the  same  arrangement  and 
ratios  among  coils  is  made  for  each  decade,  the  difference 


I    L 

L. 


SO  GO  GO  GO  i 

1000  100  10 


FIG.  118.     Diagram  of  connections  in  decade  form  of  bridge. 

being  that  one  decade  represents  1000-ohm  steps,  the 
next  100-ohm  steps,  etc.  This  gives  a  total  range  in  the 
four  decades  of  9999  ohms,  by  one-ohm  steps. 

The  ratio  arms  are  made  by  "  plugging  in  "  between  bar 
A  and  any  block  for  one  arm,  and  between  bar  B,  and  any 
other  block  for  the  other  arm. 


MEASUREMENT  OF  RESISTANCE 


123 


FIG.  119. — Wheatstone  bridge,  dial  form. 


FIG.  120. — Diagram  of  connections  in  dial  form  of  bridge. 


124  ELEMENTS  OF  ELECTRICITY 

The  arrangement  of  the  resistance  coils  which  compose 
these  arms  is  clear  from  the  diagram. 

The  "  Dial  "  form  of  bridge  is  shown  in  Fig.  119.  It  is 
seen  by  inspecting  Fig.  120,  that  the  ratio  arms  are  formed 
as  in  the  "  decade "  style.  The  balancing  resistance  is 
adjusted  by  turning  the  brushes,  by  means  of  the  knurled 
head,  from  block  to  block  on  the  dials.  The  resistance 
coils  are  connected  between  these  blocks  as  per  Fig.  120. 

By  means  of  this  form  of  bridge'  a  balance  may  be  obtained 
in  the  shortest  possible  time. 

The  galvanometers  used  in  connection  with  these  bridges, 
are  described  in  detail  in  Chapter  XIV. 


MEASUREMENT  OF  RESISTANCE  125 


SUMMARY    OF    CHAPTER   V 

MIL-FOOT.  A  round  wire  one  foot  long  and  one  circular 
mil  cross-section  area. 

A  circular  mil  is  the  area  of  a  circle  one  mil,  i.e.,  one- 
thousandth  of  an  inch,  in  diameter. 

The  area  of  a  circle  in  circular  mils  equals  the  square  of  the 
diameter  in  mils.  A  =d2. 

SPECIFIC  RESISTANCE.  The  resistance  of  a  mil-foot  of 
any  material.  (Sometimes  also  taken  as  the  resistance  of  a 
centimeter  cube,  from  face  to  face  opposite.) 

The  specific  resistance  of  commercial  copper  at  20°  C.  =  io.4 
ohms.  Symbol  =K. 

RESISTANCE  OF  WIRE.  The  resistance  of  a  wire  equals 
the  specific  resistance  (resistance  per  mil-foot)  multiplied  by 
the  length  in  feet  and  divided  by  the  section  area  in  circular 
mils. 


For  copper  at  25°  C. 

: 


TEMPERATURE  COEFFICIENT  OF  RESISTANCE.  For 
every  degree  centigrade  rise  in  temperature,  the  resistance  of 
a  pure  copper  wire  increases  .0042  of  its  resistance  at  o°  C. 
This  constant  .0042  is  called  the  Temperature  Coefficient  of 
Resistance.  It  has  a  much  lower  value  in  alloys,  and  a  nega- 
tive value  in  carbon,  porcelain,  etc.  If  coefficient  of  metals  is 
based  on  the  resistance  at  some  point  above  o°  C.  it  becomes 
smaller.  See  table.  Other  pure  metals  have  about  the  same 
coefficient  as  copper. 

RESISTANCE  OF  COPPER  AT  TEMPERATURE  (T). 
Based  on  resistance  at  o°  C., 

Rt=R0(i±.oo42T). 
Based  on  resistance  at  25°  C., 

Rt=R25(i±.oo38iT). 
Based  on  any  initial  temperature  (Tf), 

Rt=Rid±aT), 
where  T  =  change  in  temperature. 


126  ELEMENTS  OF  ELECTRICITY 

TEMPERATURE  RISE  MEASURED  BY  RESISTANCE 
RISE.  Any  one  of  the  above  equations  can  be  used  to  find  the 
temperature  rise  of  copper  wire,  when  hot  and  cold  resistances 
are  known. 

T=Rt~R0 

or 

T=.Rt-R25 


'.oo38iR25' 
or 

Rt-Ri 


T= 


aRi 


METHODS   OF   MEASURING    RESISTANCE. 

(1)  Fall  of  potential. 

(a)  Standard    resistance  and    voltmeter    which  need 

not  be  accurately  calibrated  to  read  volts. 

(b)  Accurate    method    for  measuring  low  resistance 

if  voltmeter  has  high  resistance  or  if  a  poten- 
tiometer is  used. 

(2)  Voltmeter-ammeter. 

(a)  Makes  use  of  instruments  in  general  use,   but 
which  must  be  accurately  calibrated. 

(3)  Wheatstone  bridge. 

(a)  Three  standard  variable  resistances  and  galvan- 

ometer. Has  advantage  of  being  a  "  no-deflec- 
tion "  method,  thus  galvanometer  need  not  be 
calibrated.  A  slide  wire  may  take  the  place  of 
two  of  the  resistances. 

(b)  Modifications  of  this  bridge  are  used  in  the  Murray 

and  Varley  "loop  "  methods  of  locating   faults 
in  cables. 

(4)  Voltmeter,  for  high  resistance. 

A  simple  method  of  ascertaining  whether  or  not  insu- 
lation resistance  between  different  parts  of  a  machine 
exceeds  a  certain  minimum. 

(5)  Insulation  resistance  of  covered  wire. 

Galvanometer  "  constant  "  found  by  noting  deflection 
when  a  standard  resistance  takes  the  place  of  the 
coil.  Galvanometer  deflections  must  be  propor- 
tional to  current  through  it.  Source  of  power  must 
be  of  constant  potential.  Wire  must  be  soaked  for 
24  hours. 


MEASUREMENT  OF  RESISTANCE  127 


PROBLEMS    ON  CHAPTER  V 

38-5.  A  copper  wire  is  500  ft.  long  and  .038  in.  in  diameter. 
How  many  volts  will  it  take  to  send  1.5  amperes  through  it? 

39-5.  What  will  be  the  drop  per  mile  in  a  line  wire  consisting 
of  No.  10  (B.  &  S.)  copper  carrying  32  amperes? 

40-5.  What  will  be  the  line  loss  in  voltage  and  in  watts  per 
mile  in  transmitting  10  K.W.  at  550  volts  if  a  No.  00  copper  wire 
is  used? 

41-5.  A  group  of  incandescent  lamps  takes  12  amperes.  The 
line  loss  is  not  to  exceed  1.6  volts.  What  must  be  the  size  of  the 
copper  wire  to  be  used  if  the  lamps  are  1500  ft.  from  the  generator? 

42-5.  What  size  must  line  wires  be  in  Problem  38-3  if  motor 
is  500  ft.  from  generator? 

43-5.  If  wire  used  in  line  in  Problem  42-3  is  No.  6  (B.  &  S.), 
how  far  must  lamps  be  from  generator? 

44-5.  Distance  between  M j  and  M  2  is  200  ft.,  Problem  51-3. 
What  size  is  line  wire  between  motors?  Temperature  of  wire  = 
50°  C. 

45-5.  To  what  size  wire,  B.  &  S.,  are  feeder  and  trolley,  joined 
as  in  Problem  57-3  equivalent? 

46-5.  A  rough  rule  for  the  safe  carrying  capacity  of  copper  is, 
"1000  amperes  per  sq.in.  cross-section."  According  to  this  rule 
what  should  be  the  diameter  of  a  round  wire  capable  of  carry- 
ing 250  amperes? 

47-5.  According  to  rule  in  Problem  46,  what  should  be  safe 
carrying  capacity  of  No.  0000  (B.  &  S.)? 

48-5.  At  65°  C.,  what  is  the  resistance  per  1000  ft.  of  No.  6 
copper  wire? 

49-5.  A  trolley  wire  consists  of  No.  0  hard-drawn  copper. 
What  will  be  the  drop  per  mile  onNa  day  when  the  temperature 
of  the  wire  is  —10°  C.,  if  the  wire  carries  45  amperes? 

50-5.  A  generator  is  supplying  10  amperes  to  50  arc  lamps 
in  series.  Drop  across  each  lamp  =  45  volts.  Twelve  miles  of 
No.  10  wire  are  used  for  line.  Find  brush  potential  of  generator 
if  temperature  of  wire  is  25°  C. 

51-5.  If  generator  has  65  ohms  resistance  and  line  wire  is  at 
temperature  of  —5°  C.,  what  E.M.F.  and  brush  potential  must 
be  used  in  Problem  50? 

52-6.  Compute  efficiency  of  transmission  in  Problems  50  and  51. 


128 


ELEMENTS  OF  ELECTRICITY 


53-5.  A  110- volt  system  has  an  insulation  resistance  for  each 
wire  of  200  megohms  per  mile.  What  will  the  leakage  be  on  a  5- 
mile  line? 

54-5.  Insulation  resistance  should  be  high  enough  so  thai 
not  more  than  one-millionth  of  the  rated  current  leaks  through 
the  insulation.  On  this  basis,  what  should  be  the  insulation  per 
mile  of  a  2-mile  line,  transmitting  120  K.W.  at  550  volts? 

55-5.  According  to  rule  in  Problem  46,  what  size  wire  B.  &  8. 
should  be  used  in  system  of  Problem  54? 

56-5.   In  Fig.  121  the  wires  used  are  No.  6  B.  &  8.,  copper. 
AS  =  1500  ft,; 
£C  =  #(7  =  700  ft.; 


Each  lamp  takes  2  amperes. 
Find: 

(1)  Line  loss  in  each  section. 

(2)  Voltage  across  each  group. 

(3)  *  Efficiency  of  transmission. 


ii- 


III 


FIG.  121. 

57-5.  A  line  consisting  of  copper  wire  has  a  resistance  of  2 
ohms  at  0°  F.  What  is  its  resistance  at  85°  F.? 

58-5.  It  is  generally  specified  that  the  temperature  of  the  field 
coils  of  a  dynamo  must  not  rise  more  than  50°  C.  on  full  load. 
The  resistance  of  a  set  of  field  coils  before  running  was  80  ohms 
at  20°  C.  After  run  of  3  hours  at  rated  load  the  resistance  became 
92.4  ohms.  Did  machine  meet  the  usual  specifications? 

59-5.  The  resistance  of  a  car  heater  when  cold  (20°  C.)  is  120 
ohms.  If  the  temperature  rises  to  150°  C.  when  hot,  how  much 
less  current  does  it  take  when  hot  than  when  cold?  Material  of 
heater  is  iron  wire.  Voltage  is  550. 

60-5.  It  is  desired  to  keep  the  current  constant  through  a 
coil  of  copper  wire,  which  is  on  a  constant  voltage  circuit.  The 
resistance  of  coil  at  85°  C.  is  40  ohms.  How  many  feet  of  Boker's 


MEASUREMENT  OF  RESISTANCE 


129 


IAIA  wire,  No.  18  B.  &  S.,  must  be  added  to  copper  wire  when 
the  temperature  falls  to  25°  C.  in  order  that  current  may  not  change? 

61-5.  A  motor  2000  ft.  from  the  generator  requires  50 
amperes  at  550  volts.  What  wire  should  be  used  according  to 
rule  in  Problem  46-5?  What  is  brush  potential  of  the  gene- 
rator and  the  power  lost  in  line?  Efficiency  of  transmission? 

62-5.  If  wire  of  twice  the  cross-section  is  used  in  Problem  61-5, 
what  would  answers  be? 

63-5.  A  110- volt  25-H.P.  motor  of  90  per  cent  efficiency  is 
situated  500  ft.  from  the  generator.  No.  0000  B.  &  S.  copper  wire 
is  used  for  line.  What  must  brush  potential  of  generator  be? 
Efficiency  of  transmission? 

64-5.  What  size  wire  might  have  been  used  in  Problem  63  if  a 
line  drop  of  4  per  cent  of  brush  potential  of  generator  was  desired? 

65-5.  What  size  aluminum  wire  could  have  been  used  in  Prob- 
lem 63-5  and  have  the  same  line  drop? 

66-5.  What  size  of  copper  wire  is  required  between  a  11 5- volt 
generator  and  a  110- volt  10-H.P.  motor  of  85  per  cent  efficiency? 
The  motor  and  generator  are 
800  ft.  apart. 

67-5.  In  Fig.  122,  the  num- 
bers represent  the  ohms  resis- 
tance in  that  part  of  the  circuit. 
The  battery  has  a  terminal  po- 
tential of  2  volts. 

Find: 

Current  in  AB. 

11  AD. 

"  BD. 

"        "  BC. 

"  DC. 

68-5.  A  building  situated  1200  ft.  from  a  115- volt  generator 
is  to  be  supplied  with  sufficient  current  from  the  generator  to  light 
500  lamps  each  taking  .45  ampere.  The  efficiency  of  transmission 
must  be  97  per  cent.  What  size  copper  wire  must  be  used? 

69-5.  In  Fig.  Ill,  voltmeter  has  resistance  of  15,000  ohms. 
Voltage  across  AB  =  ll2  volts.  When- connected  as  shown  in 
diagram,  voltmeter  reads  4  volts.  What  is  the  insulation  resistance 
between  armature  coils  and  frame? 


FIG.  122. 


CHAPTER  VI 
MAGNETIC    FIELD  DUE  TO  A  CURRENT 

Field  within  a  Coil — Ampere-turns — Magnetomotive  Force.  Re- 
luctance— Ohm's  Law  of  the  Magnetic  Current — Relation  between 
B  and  H — Permeability  at  Different  Degrees  of  Magnetization — 
Three  Stages  of  Magnetization — Saturation  Point — Hysteresis. 

91.  Field  within  a  Coil.  We  have  seen  in  Chapter  II 
that  wherever  there  is  an  electric  current  there  is  always 
present  a  magnetic  field.  An  electric  current,  then,  possesses 
the  property  of  producing  and  maintaining  a  magnetic 
field.  This  property  of  an  electric  current  is  called  its 
magnetomotive  force.  The  strength  of  this  field  depends 
upon  the  strength  of  the  current,  the  shape  of  its  path,  and 
the  permeability  of  the  medium  in  which  the  magnetic  field 
is  set  up. 

For  instance,  the  field  about  a  straight  wire  in  air  is  com- 
paratively weak,  while  the  field  on  the  inside  of  a  solenoid 
containing  an  iron  core  constitutes  our  most  powerful 
electromagnet. 

It  is  this  magnetic  field  of  a  solenoid  that  we  wish  to  con- 
sider here.  A  solenoid  is  a  long  coil,  or  a  coil  the  length 
of  which  is  great  in  comparison  to  its  diameter. 

It  has  been  discovered  that  if  we  wind  a  solenoid  so  that 
there  is  one  turn  of  wire  to  every  centimeter  of  the  length 
of  the  solenoid,  and  send  1  ampere  through  the  wire,  then 
the  field  intensity  in  the  air  on  the  inside  of  the  solenoid 

is  1.26  gausses  or  —  gausses.     That  is,  there  is  an  average 

of  1,26  lines  per  sq.cm.  cross-section  of  the  solenoid. 

130 


MAGNETIC  FIELD  DUfi  TO  A   CURRENT  131 

This  is  generally  written  in  the  form  of  an  equation. 
^ 


101 

where  H  =  field  strength  in  gausses', 

N  =iot&\  number  of  turns  on  solenoid; 
/=  current  in  amperes', 
I  =length  of  solenoid  in  centimeters. 

Example.  What  is  tHe  field  strength  within  a  solenoid  in  the 
shape  of  a  ring,  30  cms.  long,  consisting  of  150  turns  of  wire  carry- 
ing a  current  of  8  amperes?  Air  core. 

AT  =  150  turns. 
1=  30  cms. 
7  =  8  amperes. 

1.26.X  150X8 
H  = — =  50.4  gausses. 

Since  H,  the  number  of  lines  per  sq.cm.  set  up  in  air, 
is  also  the  measure  of  the  magnetizing  force,  it  is  possible 
to  find  the  number  of  lines  of  force  per  sq.cm.  set  up  in  any 
material,  providing  its  permeability  /*  is  known. 

For  instance,  if  the  coil  in  the  above  example  contained 
an  iron  core,  the  permeability  of  which  was  800,  the  flux 
density,  B,  in  the  iron  would  be  800X50.4  or  40,300  gausses. 
Of  course,  in  order  to  compute  the  complete  number  of  lines 
in  the  iron,  it  would  be  necessary  merely  to  know  the  cross- 
section  area  of  the  iron.  That  is,  <f>  =BA. 

This   equation,   then,   H= = ,  is  the  fundamental 

L 

equation  on  which  the  computation  of  all  magnetic  circuits 
is  based.  Its  most  important  use  is  in  the  computation 
of  the  number  of  amperes  and  turns  of  wire  necessary  to 


132  ELEMENTS  OF  ELECTRICITY 

magnetize  given  magnetic  circuits  up  to  a  certain  flux 
density. 

That  part  of  the  equation  representing  the  product, 
NI  (the  amperes  times  the  number  of  turns),  is  called  the 
AMPERE-TURNS.  As  this  expression  is  divided  by  the 
length  I,  it  is  evident  that  the  value  of  H  depends,  not 
upon  the  total  number  of  AMPERE-TURNS,  but  upon  the 
AMPERE-TURNS  per  centimeter  of  length. 

Note  then  in  particular  that  the  equation,  H=     — , 

l 

gives  the  magnetizing  force  for  one-centimeter  lengths  only. 
So  in  using  it  to  compute  the  necessary  ampere-turns  to 
magnetize  a  piece  of  iron,  remember  that  the  value  of 
NI  obtained,  will  be  sufficient  for  merely  a  centimeter  of 
iron.  If  the  iron  has  a  greater  length  than  one  centimeter 
the  result  must  be  multiplied  by  this  length. 

Problem  1-6.  What  is  the  magnetizing  force  of  a  coil  forming 
a  closed  ring  having  200  turns  to  the  centimeter  and  carrying  .04 
amperes? 

Problem  2-6.  What  would  be  the  flux  density  in  a  piece  of 
iron  placed  within  the  coil  of  Problem  1?  (/z  =  500.) 

Problem  3-6.  rfow  many  ampere-turns  per  centimeter  are 
necessary  to  produce  a  field  intensity  of  16  gausses  within  a  solenoid 
having  a  wooden  core? 

Problem  4-6.  How  many  ampere-turns  are  necessary  to  pro- 
duce a  field  strength  of  16  gausses  with  a  solenoid  40  cms.  Icng? 
Wooden  core. 

Problem  5-6.  It  is  desired  to  magnetize  a  piece  of  iron  (//  =  1000) 
to  a  flux  density  of  9000  gausses.  (a)  What  magnetizing  force 
is  necessary?  (6)  How  many  ampere-turns  per  cm.  are  required? 

Problem  6-6.  If  solenoid  used  to  magnetize  iron  in  Problem 
5-6  has  50  turns  per  cm.,  how  many  amperes  must  be  used? 

Problem  7-6.  If  iron  in  Problem  5-6  is  10  cms.  long,  how 
many  ampere-turns  are  necessary? 

Problem  8-6.  A  ring  of  iron  25  cms.  long  is  wound  by  a  coil 
of  equal  length.  What  will  be  flux  density  within  iron  if  the  coil 
consists  of  1250  turns  of  wire  carrying  .4  ampere?  (^  =  750). 


MAGNETIC  FIELD  DUE  TO  A   CURRENT  133 

Problem  9-6.  '  How  many  ampere-turns  are  needed  to  magnetize 
30  cms.  of  steel  of  a  permeability  of  1200  to  a  flux  density  of 
16,000  gausses? 

Problem  10-6.  A  ring  of  iron  24  cms.  average  length  and  1.5 
sq.cm.  cross-section,  is  wound  with  960  turns  of  wire.  If  .42 
amperes  are  sent  through  wire  what  is  flux  density  in  the  iron? 

(/i  =  850.) 

Problem  11-6.  (a)  What  is  the  total  flux  in  iron  of  Problem 
10-6?  (6)  What  is  its  magnetic  strength  in  unit  poles? 

Problem  12-6.  A  cylindrical  piece  of  iron  30  cms.  long,  3.6 
cms.  in  diameter,  is  bent  into  a  circle  and  wound  with  1200  turns 
of  wire.  How  many  amperes  must  be  sent  through  the  wire  to 
produce  a  magnet  of  24,000  unit  poles?  (/<  =  640.) 


92.  Magnetomotive  Force.  In  the  above  section  we 
have  seen  that  H,  which  equals  -—, ,  is  the  force  tending 

to  magnetize  one  centimeter  length  of  core,  and  that  if  we 
wish  to  magnetize  a  length  /  of  more  than  one  centimeter 
we  must  multiply  the  above  force  by  the  length  /.  This 
expression,  HI,  is  sometimes  called  the  MAGNETOMOTIVE 
FORCE  symbol,  M. 

This  magnetomotive  force  M  might  be  denned  as  the 
PRESSURE  which  forces  the  magnetic  flux  or  current  through 
the  circuit,  and  corresponds  to  the  electromotive  force  E, 
or  voltage  which  forces  the  electric  current  through  its 
circuit.  They  both  are  pressure;  M  is  magnetic  pressure, 
and  E  is  electric  pressure. 

We  have  said  that  the  magnetic  pressure  M  of  a  long  coil 
equals  the  field  strength  times  the  field  length.  That  is, 
adding  to  the  length  /  of  the  magnetic  path,  adds  to  the 
pressure  needed  to  set  up  a  given  flux  density  throughout 
the  circuit. 

Likewise  adding  to  the  flux  density  to  be  set  up  throughout 
the  circuit  adds  to  the  pressure  needed  to  set  up  the 
flux. 


134  ELEMENTS  OF  ELECTRICITY 

Thus,  as  we  have  seen; 

M=Hl. 

„     1.26A7 
But  H= — - — . 

Then  M  =  i- 

This  may  be  written, 


Of  course  if  we  are  considering  one  centimeter  length  of 
coil,  the  field  intensity  H  equals  the  magnetomotive  force  M. 
93.  Comparison  of  Equations  for  H  and  M.      We  have 
then,  these  two  very  similar  equations,  and  the  distinction 
between  them  must  always  be  kept  clearly  in  mind. 
The  one: 

1.26JV7 
/      ' 

determines  the  field  strength  H  in  lines  per  sq.cm  within 
a  long  coil  with  an  air  core,  or  the  force  necessary  to  magnet- 
ize one  centimeter  length  of  the  air  core. 
The  other: 


determines  the  magnetomotive  force  (pressure)  tending  to 
set  up  the  magnetic  field  throughout  the  entire  path  of  any 
kind  of  material. 

The  equation  for  M  is  rarely  used,  'it  being  customary 
to  compute  the  magnetizing  force,  H,  for  one  centimeter 

1  2QNI 
length  by  the  equation  H  =-^— ,  and   then   to   multiply 

the  result  by  the  length  of  the  path  (I) ,  as  in  previous  prob- 
lems in  this  chapter. 

However,  the  term  magnetomotive  force,  M,  is  used  when 
problems  in  magnetism  are  solved  by  means  of  what  may 


MAGNETIC  FIELD  DUE  TO  A  CURRENT  135 

be  called  " Ohm's  Law  of  the  Magnetic  Circuit"  as  explained 
below. 

94.  Reluctance.  Ohm's  Law  of  the  Magnetic  Circuit. 
The  analogy  between  a  magnetic  circuit  and  an  electric 
circuit  is  so  complete  that  Ohm's  law  may  be  applied  to 
both. 

It  has  just  been  shown  that  there  is  a  magnetic  pressure 
M  which  causes  a  magnetic  current  to  flow  in  a  magnetic 
circuit,  exactly  as  an  electric  pressure  (voltage)  E  causes 
an  electric  current  to  flow  in  an  electric  circuit. 

There  is  a  magnetic  current  consisting  of  the  lines  of 
force  (j>  just  as  there  is  an  electric  current  consisting  of  the 
amperes  /. 

Similarly,  there  is  a  magnetic  resistance  in  a  magnetic 
circuit  which  opposes  the  flow  of  the  magnetic  current, 
just  as  the  electric  resistance  opposes  the  flow  of  an  electric 
current.  This  magnetic  resistance  is  called  RELUCTANCE, 
(R,  in  order  to  distinguish  it  from  electric  RESISTANCE,  R. 
There  is  no  recognized  name  for  a  unit  of  reluctance. 

The  electric  current  .flowing  through  a  circuit  equals  the 
electromotive  force  divided  by  the  resistance. 

The  magnetic  current  flowing  through  a  magnetic  cir- 
cuit equals  the  magnetomotive  force  divided  by  the  reluc- 
tance. By  equations: 

7=f     (Electric); 
n 

<£=—     (Magnetic); 

/  =  electric,  current  in  Amperes. 
(/)  =magnetic  current  in  Lines. 

E  =  electric  pressure  in  Volts. 
M  =magnetic  pressure  in  (Gilberts). 

R  =electric  resistance  in  Ohms. 
(R=magnetic  reluctance  in  (Oersteds). 


136  ELEMENTS  OF  ELECTRICITY 

Thus  in  order  to  find  the  number  of  lines,  </>,  threading 
a  magnetic  circuit,  it  is  necessary  to  find  the  magnetic 
pressure  M  and  divide  it  by  the  reluctance  (R. 

Example.  The  magnetic  circuit  of  an  electro -magnet  has  a 
reluctance  of  0.002.  How  many  lines  of  force  thread  the  circuit 
if  the  magnetomotive  force  is  16  units. 

By  Ohm's  law  of  magnetic  circuits : 


.002' 
=  8000  lines. 


Example.     How  many  ampere-turns  are  necessary  to  supply 
the  magneto-motive  force  in  above  example? 


vr-  M 


1.26 

J_6 
~1~26 

=  12.7  ampere-turns. 

Problem  13-6.  The  magnetic  circuit  of  an  electro-magnet  has 
0.4  unit  of  reluctance.  How  great  a  magnetomotive  force  is 
necessary  to  set  up  a  flux  100,000  lines? 

Problem  14-6.  How  many  ampere-turns  are  necessary  in 
Problem  13-6? 

Problem  15-6.  A  certain  electro-magnet  has  a  flux  of  200,000 
lines.  The  coil  has  450  turns  carrying  1.4  amperes.  What  is 
the  reluctance  of  the  magnetic  circuit? 

Problem  16-6.  A  coil  consists  of  980  turns.  The  reluctance 
of  the  magnetic  circuit  is  .008  unit.  How  many  amperes  must 
flow  through  coil  to  set  up  a  flux  of  1,000,000  lines? 


MAGNETIC  FIELD  DUE   TO  A  CURRENT  137 

95.  Computation  of  Reluctance.  In  finding  the  reluc- 
tance of  a  magnetic  circuit  of  known  size  and  materials, 
we  use  the  same  method  employed  in  finding  the  resistance 
of  an  electric  circuit  of  known  size  and  materials. 

To  find  the  resistance  of  a  conductor  we  use  the  equation, 

R,      Kl 


d2  (area) ' 

where  K  =resistance  of  unit  (mil-ft.)  ; 

I  =length  in  feet ; 
d2  =section  area  in  circular  mils. 

Similarly,  to  find  the  reluctance  of  a  magnetic  conductor 
we  use  the  equation: 

Kl 


(R 


A  (area)' 


where  K  = reluctance  of  unit  (cubic  centimeter) ; 

Z=length  in  centimeters; 
A  =  section  area  in  square  centimeters. 

The  magnetic  tables  or  data,  however,  instead  of  giving 
K,  the  RELUCTANCE  of  a  cubic  centimeter,  generally  give 
fjL,  the  PERMEABILITY,  which  is  merely  the  inverse  of  the 
reluctance  of  a  cubic  centimeter.  This  is  exactly  analogous 
to  the  fact  that  the  electric  conductivity  is  the  inverse 
of  the  resistance  of  a  mil-ft. 

We  may  write,  then,  the  equation: 

JfJk 

ft 

In  order  to  use  the  tables,  we  generally  use  (  — J  instead  of 
(K)  and  write  the  equation  for  reluctance : 


138 


ELEMENTS  OF  ELECTRICITY 


Example.  A  piece  of  iron  is  400  cms.  long  and  has  a  cross- 
section  area  of  200  sq.cms.  Permeability  =  1000.  What  is  the 
reluctance? 

Since  the  permeability   (,«)=1000,    the   reluctance   of  a  cubic 

centimeter  CK)=  Thus 


400 


1000     200 
=  .002  units. 

Reluctances  in  series  are  added  like  resistances  in  series. 

Reluctances  in  parallel  are  treated  like  resistances  in  parallel. 

Example.  A  magnetic  circuit,  Fig.  123,  is  made  up  of  a  curved 
bar  of  iron  400  cms.  long,  cross-section  80  sq.cms.,  and  an  air  gap 
2  cms.  long.  What  is  reluctance  of  circuit?  Permeability  of 
the  iron  =  1000. 

For  the  iron 


(R  = 


for  the  air 


total  reluctance 


FIG.  123. 


(R  =  .025  +  .005  =  .03  unit. 


Problem  17-6.  What  is  the  reluctance  of  a  piece  of  cast  iron 
30  cms.  long  with  a  cross-section  of  2 X4  cms.?  (/x  =  200.) 

Problem  18-6.  A  wrought-iron  ring,  2.4  sq.  cms.  cross-section, 
has  a  mean  diameter  of  20  cms.  What  is  the  reluctance  if  the 
permeability  of  the  iron  is  900? 

Problem  19-6.  A^  magnetic  circuit  is  made  up  of  the  following 
parts  in  series:  (1st)  18  cms.  of  2X3  cms.  cast  iron,  /*  =  150;  (2d) 
25  cms.  of  3X2.5  annealed  steel,  /*  =  1400;  (3d)  2  air  gaps  each 
3.8  cms.  long  and  3.2X2.0  cms.  cross-section.  Find  reluctance 
of  circuit. 


MAGNETIC  FIELD  DUE  TO  A  CURRENT  139 

Such  problems  as  the  following  can  be  solved,  when  the  com- 
putation of  reluctance  is  possible. 

Example.  How  many  ampere-turns  are  required  to  set  up 
500,000  lines  in  an  iron  ring  250  cms.  long  and  50  sq.cms.  section 
area?  Permeability  =  500. 


<£=  500,  000  lines; 
|       |  500,000  =  !^-; 

V        NI  =  3980  ampere-turns. 

But  note  that  this  same  example  can  also  be  solved  as  follows 
by  the  same  method  used  in  the  first  twelve  problems  in  this 
chapter,  which  is  the  method  generally  used. 

Lines  per  sq.cm.  (B)  in  the  iron  must  equal  total  lines  (</>)  ' 
divided  by  area  (A)  of  ring. 


_  500,000 
50 

=  10,000  lines; 
B 

ti  =  -T7> 


H-* 


^10,000 
500 

=  20; 


140  ELEMENTS  OF  ELECTRICITY 

U_\.2QN1 

~T~; 

HI 
~1.26 

_  20X250 
1.26 

=  3980  ampere-turns. 

Problem  20-6.  How  many  ampere-turns  are  necessary  to 
magnetize  an  iron  ring  40  cms.  long,  2.8  sq.cms.  cross-section  area, 
so  that  the  total  flux  equals  50,000  lines?  (//=950.) 

Problem  21-6.  A  magnetic  circuit  consists  of  three  parts: 
(1st)  28  cms.  of  cast-iron,  5X1.5  cms.  section,  /*  =  100;  (2d)  80 
cms.  of  wrought  iron,  4X2  cms.  section,  /*=-1200;  (3d)  1.5  cms.  air 
gap,  4.5X3  cms.  section.  Find  ampere-turns  necessary  to  set 
up  100,000  lines  in  this  circuit. 

Problem  22-6.  A  flux  of  40,000  lines  is  set  up  in  magnetic 
circuit  of  Problem  19-6  by  a  coil  consisting  of  15,000  turns.  What 
current  flows  in  coil? 

Problem  23-6.  How  many  ampere-turns  are  Jiecessary  to  set 
up  a  flux  density  of  10,000  gausses  in  ring  of  Problem  18-6? 

96.  Flux  Density  (B).  Field  Intensity  .  (H).  Permea- 
bility (fi).  We  have  seen  that  by  means  of  a  form  of  Ohm's 
law,  we  can  find  the  total  flux  (0)  threading  a  magnetic 
circuit,  and  that  the  number  of  lines  per  square  centimeter 
of  any  part  of  the  circuit  is  the  FLUX  DENSITY  of  that 
part.  We  use  the  symbol  B  to  denote  flux  density  in 
any  material  except  air.  For  instance,  a  20-sq.cm.  cross- 
section  iron  circuit  might  have  a  total  flux  <£>  of  100,000 
lines.  The  flux  density  would  be  ao%jj00  or  5000  gausses. 

There  are  more  lines  set  up  in  iron  than  in  air  by  the  same 
magnetizing  force,  H,  because  the  permeability  of  iron  is 
larger.  In  fact,  the  number  of  lines  per  sq.cm.,  5,  is  as 
much  greater  in  iron  than  in  air,  H,  as  the  permeability 
of  the  iron  is  greater  than  the  permeability  of  the  air. 


MAGNETIC  FIELD  DUE  TO  A   CURRENT  141 

Thus  the  permeability  of  a  material  may  be  found  by  com- 
paring the  flux  density  produced  in  it  with  the  field  intensity 
produced  in  air.  The  field  intensity  in  air,  H,  is  also  called 
the  magnetizing  force.  We  often  define  permeability  of 
a  material,  therefore,  as  the  ratio  of  flux  density  B  set  up 
in  the  material  to  the  magnetizing  force  H.  That  is, 
the  permeability  of  a  material  is  the  ratio  of  the  number 
of  lines  of  force  per  sq.cm.,  set  up  in  that  material,  to  the 
number  that  would  be  set  up  in  air  under  the  same  con- 
ditions. 

For  instance:  A  coil  may  have  a  field  intensity,  H,  of 
10  lines  per  sq.cm.  when  the  inside  of  the  coil  is  air.  If 
the  inside  of  the  coil  were  an  iron  core,  there  might  be  40000 
lines  per  sq.cm.,  B,  in  the  iron.  There  are  then  I^LO. 
or  4000  times  as  many  lines  per  sq.cm.  set  up  in  the  iron 
as  in  the  air  under  the  same  conditions.  The  ratio  of  B 
to  H  is  4000,  or  in  other  words  the  iron  has  a  permeability 
of  4000.  It  is  often  expressed  in  the  form  of  an  equation, 
as  we  have  seen, 

B 


This  term  H,  which  we  know  represents  the  field  intensity 
in  air,  or  the  number  of  lines  of  force  per  sq.cm.  in  air, 
is  often  spoken  of  as  the  MAGNETIZING^ORCE.  This  is 
not  to  be  confused  with  M  the  magnetomotive  force 
which  is  HXl. 

97.  Permeability  Depends  upon  Degree  of  Magnetization. 
Magnetic  Permeability  has  been  compared  to  Electric 
Conductivity.  There  is,  however  this  difference:  The 
conductivity  of  a  given  piece  of  copper,  for  instance,  is 
constant,  unless  the  temperature  changes.  We  know  that 
if  we  wish  to  send  double  the  current  through  a  wire  we 
can  count  on  the  conductivity  remaining  the  same  as  though 
we  were  not  increasing  the  current. 


142 


ELEMENTS  OF  ELECTRICITY 


On  the  other  hand,  if  we  wish  to  double  the  number  of 
lines  in  a  piece  of  iron,  we  find  that  we  cannot  do  so  by 
merely  doubling  the  magnetizing  force.  Sometimes  we 
need  but  increase  the  current  a  very  little.  Again,  we 
must  use  many  times  the  original  current.  In  other  words, 
the  permeability  of  iron  or  steel  depends  upon  how  many 
lines  it  already  contains.  Another  way  of  stating  it  is: 
the  value  of  JJL  depends  upon  B,  the  magnetic  condition  of 
the  iron.  After  a  certain  point  is  reached,  as  the  flux 
density  increases,  the  permeability  decreases  very  rapidly, 
and  it  becomes  difficult  to  set  up  a  greater  number  of  lines. 

98.  Saturation.  This  makes  it  impracticable  to  magnetize 
a  piece  of  iron  beyond  a  certain  flux  density,  called  the 
SATURATION  POINT.  Of  course  it  is  possible  to  magnetize 
iron  beyond  the  saturation  point,  but  for  every  small  increase 
in  flux  density  there  must  be  a  larger  and  larger  increase 
in  magnetizing  force,  which  it  is  unprofitable  to  maintain. 

This  relation  of  B  to  H  is  clearly  shown  in  curves  plotted 
from  some  data  taken  by  students  at  Pratt,  on  a  sample 
piece  of  wrought  iron. 

WROUGHT   IRON 


B 

H 

ft 

B 

// 

/( 

1,000 

.48 

2,080 

9,000 

2.95 

3,050 

2,000 

.61 

3,280 

10,000 

4.32 

2,310 

3,000 

.78 

3,850 

11,000 

6.70 

1,640 

4,000 

.92 

4,340 

11,500 

9.46 

1,220 

6,000 

1.20 

5,000 

12,000 

12.40 

953 

7,000 

1.40 

5,000 

12,500 

16.00 

781 

8,000 

2.00 

4,000 

13,000 

23.80 

546 

In  Fig.  124,  the  Magnetization  Curve,  as  the  curve  show- 
ing the  relation  of  B  to  H  is  called,  the  point  x  denotes  the 
"  knee  "  of  the  curve  and  is  the  point  generally  spoken  of 
as  the  SATURATION  POINT.  It  is  the  point  where  a  small 


MAGNETIC  FIELD  DUE  TO  A   CURRENT 


143 


increase  of  flux  density  B  requires  a  large  increase  in  magnet- 
izing force  H.  In  this  specimen  of  iron,  it  occurs  when  a 
flux  density  of  about  -8000  lines  per  sq.cm.  has  been  set 
up.  This  requires  a  magnetizing  force  of  2.00  gausses. 

If  the  field  cores,  for  instance,  of  a  motor  were  made  of 
this  iron,  the  flux  density  at  which  it  could  be  most  profit- 
ably run,  would  be  about  8000  lines  per  sq.cm.  The  number 
of  ampere-turns  to  set  up  this  flux  density  in  each  centi- 
meter of  the  length  of  the  magnetic  circuit  of  the  motor 
can  be  determined,  as  we  have  seen.  Of  course  there  must 


12000 


10000 


8000 


O  GOOO 


co  4000 


2000 


6         7         8         9        10       11       12 
H   in  Gausses 
FIG.   124. — Magnetization  curve  of  soft  iron. 

be  added  to  this  enough  turns  to  overcome  the  reluctance 
of  whatever  air  gaps,  etc.,  there  may  be  in  the  magnetic 
circuit. 

The  principal  point  to  remember  about  the  magnetization 
of  the  iron  in  any  machine,  is  that  there  is  a  certain  flux 
density  called  the  saturation  point,  beyond  which  it  is 
possible,  but  not  practicable,  to  carry  the  magnetization.  The 
reason  for  this  is  that  it  requires  an  ever-increasing  propor- 
tion of  power  to  magnetize  iron  or  steel  beyond  this  point, 
because  of  the  rapid  decrease  in  the  permeability.1 

1  Physicists  are  accustomed  to  call  the  curve  described  above  the 
"  Induction  Curve  "  and  to  draw  as  a  Magnetization  Curve,  a, 


144  ELEMENTS  OF  ELECTRICITY 

99.  Three  Stages   of  Magnetization.       Three  stages  are 
generally  noted  in  the  magnetization  of  a  piece  of  iron. 
If  we  divide  up  the  curve,  Fig.  124,  into  the  sections  ab, 
bx,  and  xc,  the  three  stages  will  be  fairly  well  represented. 

First  Stage.  When  there  are  very  few,  or  no,  lines  in 
the  specimen  as  represented  by  the  part  ab,  it  seems 
to  be  quite  difficult  to  set  up  lines,  and  the  flux  density 
B  is  almost  proportional  to  the  magnetizing  force  H.  Ac- 
cording to  the  Magnetic  Molecule  Theory,  this  means  that 
some  difficulty  is  experienced  in  turning  the  first  few  mole- 
cules around  into  the  magnetic  position. 

Second  Stage.  But  once  these  molecules  are  turned, 
most  of  the  others  seem  to  turn  more  readily,  as  is  evidenced 
by  the  great  increase  in  flux  density  caused  by  a  slight  in- 
crease in  the  magnetizing  force.  This  is  shown  by  the 
section  of  the  curve  bx  rising  very  abruptly. 

Third  Stage.  When,  however,  the  saturation  point  is 
reached,  it  is  extremely  difficult  to  turn  the  rest  of  the 
molecules  into  the  magnetic  position,  and  there  is  required 
an  ever-increasing  magnetizing  force.  The  curve,  therefore, 
falls  off  rapidly  along  the  section  xc. 

100.  Relation  of  Permeability  to  Flux   Density.     If  we 
plot  a  curve,  as  in  Fig.  125,  between  permeability   /*  and 
flux  density  B,  we  see  that  the  permeability  starts  small  and 
increases  as  the  flux  density  increases,  until  the  latter  has 
reached   a  value    between    6000  and  7000  gausses.     Then 
suddenly,   the   permeability   begins    to    decrease    rapidly, 

curve  between  the  magnetizing  force  and  the  poles  set  up  in  the  iron. 
This  differs  but  very  little  from  the  one  used  here,  except  in  the  por- 
tion which  we  have  called  xc.  As  this  part  of  the  curve  is  never  used 
in  practical  construction,  we  have  followed  the  general  practice  in 
calling  the  above  "  BH  curve  "  the  "  MAGNETIZATION  CURVE." 

The  term  "  Saturation  Point,"  as  used  above,  indicates  the  PRAC- 
TICAL SATURATION  POINT,  and  not  the  THEORETICAL,  which,  of  course, 
is  much  further  out  on  the  curve. 

For  a  theoretical  discussion  of  this  subject  see  any  book  on  the 
theory  of  magnetization. 


MAGNETIC  FIELD  DUE  TO  A   CURRENT 


145 


which  it  would  continue  to  do  till  it  reached  the  value  of 
p  for  air. 

Thus,  in  determining  the  value  of  /*  for  iron,  we  must 
know  not  only  the  quality  of  the  iron,  but  also  the  flux 
density  at  which  it  is  to  be  used.  As  stated  previously, 
it  is  in  this  particular  that  magnetic  permeability  differs 


(jtt)  Permeability 

/ 

\ 

/ 

\ 

/ 

V 

/ 

] 

/ 

\ 

/ 

\ 

1 

\ 

/ 

/ 

\ 

/ 

\ 

I 

\ 

1 

\ 

\ 

\ 

RELATION  OF 
PERMEABILITY  TO 
FLUX  DENSITY 

\ 

\ 

\ 

\ 

2000     400U 


6000      8000     10000    12000 
B    in  Gausses 


FIG.  125. — Relation  of  permeability  to  flux  density  in  iron. 

from  electric  conductivity.  In  electrical  circuits,  it  makes 
no  difference  whether  the  current  density  is  1  ampere  or 
100  amperes,  the  conductivity  remains  the  same  (barring 
temperature  changes).  But,  as  has  just  been  shown, 
the  magnetic  permeability  changes  every  time  the  current 
(or  flux  density)  changes. 
In  using  any  equation  containing  the  quantity  //,  it  is 


146  ELEMENTS  OF  ELECTRICITY 

necessary,  therefore,  to  select  the  value  to  be  applied  to  it, 
with  due  regard  to  the  magnetic  conditions  of  the  material. 

101.  Practical  Computation  of  Ampere-turns.  It  is  on 
account  of  this  change  in  permeability,  that  in  computing 
the  ampere-turns  necessary  to  produce  a  certain  flux  in  a 
magnetic  circuit,  it  is  not  customary  to  use  the  equation 

<p=  —  .     Instead,  the  flux  density  (B)  is  computed  (5=^-), 
(R  \        A/ 

and  the  value  of  H  to  produce  this  density  is  found  in  a 
table.  The  number  of  ampere-turns  to  produce  this  value 
of  H  is  then  computed  from  the  equation 

„.     1.2QNI  A77_  HI 

I      '  ~1.26' 

as  explained  earlier  in  this  chapter. 

Example.  How  many  ampere-turns  are  required  to  set  up  a 
flux  of  1,200,000  lines  in  a  cast-iron  circuit  200  cms.  long,  with 
a  cross-section  area  of  300  sq.cms.? 


4000  gausses. 


From  table,  in  Appendix,  for  relation  of  B  to  H  in  cast-iron, 
H  (for  B  of  4000)  -8.5; 

o  c  \/  200 
NI  =  =  1350  ampere-turns. 

The  design  of  an  electro-magnet  includes  many  details  in  addi- 
tion to  the  number  of  ampere-turns.  The  size  of  wire  and  the 
amount  of  surface  required  for  radiation  are  two  factors  which 
affect  the  design  fully  as  much  as  the  ampere-turns.1 

1  For  practical  directions  in  the  design  of  electro-magnets,  the  Varley 
Electro-Magnet  Co.  of  Jersey  City  has  published  a  most  excellent  book- 
let. It  contains  a  mass  of  information  and  practical  formulas  which 
will  be  found  invaluable  to  one  designing  a  magnet  for  a  definite  use, 


MAGNETIC  FIELD  DUE  TO  A   CURRENT  147 

Solve  the  following  problems  by  use  of  BH  tables  in  Ap- 
pendix. 

Problem  24-6.  A  wrought-iron  magnetic  circuit  has  a  length  of 
65  cms.  How  many  turns  must  be  wound  on  it  to  give  a  flux  den- 
sity of  6000  gausses  if  the  current  through  the  turns  is  .75  ampere? 

Problem  25-6.  If  circuit  in  Problem  24  has  section  area  of  8  sq. 
cms.,  what  total  flux  is  set  up? 

Problem  26-6.  A  magnetic  circuit  consists  of  wrought-iron 
ring  and  air  gap  of  the  dimensions  in  Fig.  123.  It  is  desired  to 
set  up  a  flux  of  1,280,000  lines  in  this  circuit.  How  many  ampere- 
turns  must  be  wound  upon  it? 

Problem  27-6.  If  ring  in  Problem  26  were  cast-iron,  how  many 
ampere-turns  would  be  necessary  to  set  up  1,000,000  lines  in  circuit? 

Problem  28-6.  In  generator  of  Fig.  149,  how  many  amperes 
must  be  sent  through  field  coils  containing  400  turns  each  to  set  up 
1,200,000  lines  per  pole?  Poles  and  yoke  are  of  wrought  iron. 
Poles  40  cms.  long,  area  100  sq.  cms.  Path  of  flux  through  yoke 
from  pole  to  pole,  100  cms.  long,  80  sq.  cms.  area.  Consider  each 
magnetic  path  through  the  armature  10  cms.  long,  120  sq.  cms. 
area.  Air  gap  .8  cm.  long,  130  sq.  cms.  area. 

102.  Hysteresis.  In  Fig.  124,  we  saw  how  the  flux 
density  B  increases  when  the  magnetizing  force  H  increases, 
even  when  the  curve  was  carried  to  a  point  considerably 
beyond  saturation.  If  now  we  gradually  take  away  the 
magnetizing  force,  let  us  see  what  happens. 

In  Fig.  126,  the  line  abxc  is  the  magnetization  curve 
of  Fig.  124.  At  point  Cthe  flux  density  is  11,800  gausses 
and  the  magnetizing  force  12.5  gausses. 

When,  however,  the  magnetizing  force  H  is  decreased, 
the  flux  density  B  does  not  decrease  along  the  same  line 
by  which  it  had  increased,  but  follows  the  curve  cdef.  At 
the  point  e,  although  the  magnetizing  force  H  has  become 
zero,  there  still  remains  a  flux  density,  B,  of  8800  gausses. 
The  flux  B,  therefore,  is  said  to  LAG  behind  the  force  H. 
This  lagging  is  called  HYSTERESIS,  and  is  the  cause  of  a 
certain  loss  in  every  alternating  current  machine  and  in 
the  armatures  of  D.C.  machines.  The  reason  for  this 
is  seen  if  we  consider  the  rest  of  the  curve. 


148 


ELEMENTS  OF  ELECTRICITY 


At  the  point  e,  as  has  been  said,  although  H  has  become 
zero,  the  value  of  B  has  only  decreased  to  8800,  represented 
by  the  line  ae.  This  value  is  called  the  REMANENCE.  To 
get  this  magnetic  Remanence  out  of  the  iron,  it  is  necessary 
to  set  up  a  magnetizing  force  H  in  the  opposite  direction,  of 
1.2  gausses.  This  force  is  represented  by  the  line  a/,  and 
is  called  the  COERCIVE  FORCE. 


10     8     6     4      2     0     2     4      6      8    10    12 
FIG.  126. — Hysteresis  loop  for  soft  iron. 

If  now  this  magnetizing  force  in  the  opposite  direction, 
—  H,  is  increased,  a  flux,  —  B,  will  be  set  up  in  the  opposite 
direction  and  a  curve  of  magnetization  in  this  direction 
can  be  drawn  as  fg,  until  B  has  as  large  a  negative 
value  at  g  as  it  had  a  positive  value  at  c. 

The  magnetizing  force  is  again  gradually  decreased,  and 
again  the  flux  "  lags  "  behind  the  force,  until  when  H  has 
again  become  zero,  the  value  of  B  has  decreased  to  i  only. 
There  is  thus  a  remanence  of  about  9000  gausses,  ai,  left 


MAGNETIC  FIELD  DUE   TO  A   CURRENT 


149 


in  this  direction,  to  remove  which  again  requires  a  coercive 
force  ak. 

To  bring  the  flux  density  up  to  the  value  c  again,  we 
merely  have  to  increase  this  force  to  its  original  value 
for  density  at  this  point.  The  cycle  thus  completed  is 
called  the  "  Hysteresis  Loop." 

It  can  be  shown  that  the  area  of  this  loop  represents 
the  work  done  in  overcoming  the  Hysteresis  effect  on  a 
cubic  centimeter  of  this  specimen  of  iron,  just  as  the  area 


HYST 


ER 


SIS  LOi 


FIG.  127. — Hysteresis  loss  in 
annealed  steel. 


FIG.  128  — Hysteresis  loss  in 
hard  steel. 


of  a  steam  indicator  card  represents  the  work  done  in 
overcoming  the  resistance  to  motion  offered  by  the  piston. 
Only,  in  this  case,  all  work  done,  as  represented  by  the 
area  of  the  loop,  is  wasted. 

This  can  be  shown  as  follows : 

Since  the  value  of  B,  ie.,  the  magnetic  density  of  the  iron,  always 
lags  a  little  behind  the  magnetizing  force,  some  extra  force  must 
be  exerted  continually  to  urge  on  the  change  in  the  magnetization. 
When  the  flux  density  is  being  increased,  the  value  of  H,  or 
magnetizing  force,  has  to  be  a  little  greater  than  would  be  neces- 
sary, were  there  no  hysteresis  effect.  Similarly,  as  the  iron  is 


150  ELEMENTS  OF  ELECTRICITY 

being  demagnetized,  the  magnetizing  force  must  be  larger  in  the 
opposite  direction  in  order  to  overcome  the  drag  of  the  magnetic 
lines  which  seem  to  persist  in  the  iron.  Thus  there  is  a  constant 
action  taking  place  between  the  magnetized  iron  and  the  mag- 
netic field  of  the  coil. 

This  action  between  a  magnet  and  a  magnetic  field  has  been 
described  in  Chapter  I  and  has  been  shown  to  be  equal  to  the 
product  of  the  magnetic  strength  of  the  iron  (ra)  times  the  strength 
of  the  field  H. 

In  the  form  of  an  equation  this  becomes 


where  F=iorce  in  dynes; 

m  =  strength  of  magnet  in  unit  poles; 
ft  ==  strength  of  field  in-gausses.  /  "/i^ 

The  cubic   centimeter  of  iron,   which  we   are   considering,   is 

n 

magnetized  to  a  density  of  B  gausses  and  must  contain  —  unit 

poles.     The  action  of  the  field  H  of  coil  upon  the  iron  must, 
therefore,  be  represented  by  the  equation, 


47T 
BH  j 

=—  dynes. 

Inasmuch  as  we  are  dealing  with  but  one  cubic  centimeter  of  iron, 

fiJ-f 
this  force.  --  dynes,  must  be  acting  through  a  centimeter  length 

4?r 

only.     That  is,  the  magnetization  is  made  to  change  throughout 
one  centimeter  length. 

Now,  when  a  force  of  one  dyne  acts  through  a  centimeter  space,  it 

ftJ-f 
does  one  erg  of  work.     Thus,  -  -  dynes,  acting  through  one  centi- 

T)TT 

meter,  would  do  ——  ergs  of  work. 

TtTT 

}3ut  the  value  of  -  is  continually  changing  during  the  cycle  of 

magnetization. 

The  total  amount  of  work  done  during  one  cycle  would  then  be 

BH 

the  summation  of  all  the  different  values  of  —  —  throughout  the 

47T 


MAGNETIC  FIELD  DUE  TO  A  CURRENT          151 

cycle.  The  area  of  the  loop,  in  BH  values,  represents  this  summa- 
tion, and  accordingly,  the  amount  of  energy  consumed  in  over- 
coming the  magnetic  lag,  or  hysteresis  during  one  cycle. 

It  is  customary  to  compute  this  energy  loss  in  watt-sees.  ;  1  watt- 
sec.  =107  ergs. 

The  equation  for  hysteresis  loss  in  1  cu.cm.  thus  becomes 

area  of  loop  (BH  units) 

ly  =  - 
47TX107 

Example  1.  Find  the  hysteresis  loss  per  cubic  centimeter,  per 
cycle,  in  the  iron  used  to  obtain  loop  in  Fig.  126.  Area  of  loop 
equals  24  sq.in. 

Solution.     On  horizontal  axis: 

1  in.  =  2  gausses;  magnetizing  force  H. 
On  vertical  axis  : 

1  in.  =  2000  gausses;  flux  density  B. 
1  sq.in.  -2X2000  -4000  (BH  units). 

Area  of  curve  : 

24  sq.in.  -24X4000  =  96000  (BH  units). 

Hysteresis  loss: 

96000  (BH  units) 

47TX107 

=  0.000764  watt-sees. 

Example  2.  If  20  Ibs.  of  the  iron  in  Example  1  go  through 
the  cycle  of  magnetization  60  times  per  second,  what  power  would 
be  lost  on  account  of  hysteresis? 

Solution. 

20  Ibs.  =   OX454 


From  above  example, 

1  cu.crn.  loses  0.000764  watt-sec,  per  cycle. 
Then  1230  cu.cm.  would  lose 

0.000764X60X1230  watt-sec,  in  60  cycles 

=  5.5  watt-sees. 

Since  this  loss  occurred  in  one  second  the  power  lost 
=  5.5  watts. 

With   every   cycle    (double   alternation)    of    an    electric 
current  in  a  machine,  there  is  always  this  hysteresis  loss 


152  ELEMENTS  OF  ELECTRICITY 

in  the  iron  cores.  The  greater  the  number  of  alternations 
a  second,  the  greater  the  total  loss.  For  this  reason,  the 
iron  to  be  used  in  Alternating  Current  machines,  and  in 
armatures  in  general,  has  to  be  selected  with  proper  care 
as  to  its  hysteresis  qualities. 

If  a  piece  of  annealed  wrought  iron  or  annealed  silicon- 
steel  is  put  to  this  test,  the  loops  plotted  will  be  very  narrow 
and  contain  little  area  (Fig.  127),  showing  the  loss  to  be 
small. 

In  the  case  of  hard  steel,  the  loop  will  widen  out  (Fig.  128), 
and  contain  a  large  area.  For  this  reason,  either  annealed 
iron  or  annealed  silicon-steel  is  used  in  transformer  cores, 
armatures,  etc. 

103.  Computation  of  Hysteresis  Loss.  Dr.  Steinmetz 
has  given  the  following  equation  for  computing  the  hys- 
teresis loss: 


107 

where     P=loss  per  cu.cm.  of  iron  for  1   cycle  per  sec.  in 

watts. 

Bmax  ^maximum  flux  density  in  gausses. 
K=  constant,  depending  on  material. 

The  following  table  of  the    values  of   K  is   taken  from 
Foster's  "  Electrical  Engineer's  Pocket  Book." 

HYSTERESIS  CONSTANTS  FOR  DIFFERENT  MATERIALS 

Hysteresis 

Material.  Constant. 

K 

Best  annealed  transformer  sheet  metal 001 

Very  soft  iron  wire 002 

Thin  good  sheet  iron 003 

Thick  sheet  iron 0033 

Most  ordinary  sheet  iron 004 

Transformer  cores 003 

Soft  annealed  cast  steel 008 

Soft  machine  steel 0094 

Cast  steel 012 

Cast  iron 016 

Hardened  cast  steel . .  .025 


MAGNETIC  FIELD  DUE  TO  A  CURRENT  153 

Example.  What  loss  in  power  is  due  to  hysteresis  in  a  machine 
that  contains  2500  cu.cms.  of  annealed  iron,  which  is  subjected 
to  60  cycles  a  second?  K  =  .003.  Maximum  Flux  Density  =  12,000. 


p= 


107 

.OOSXl^OOO1'6 
107 

.003X3,370,000 


107 
=  .00101  watt  per  cu.cm.  at  1  cycle  per  sec. 

Total  p^wer  lost  =  .00101  X  60  X  2500 
=  152  watts. 


,al  powei 


Example.  If  cast  iron,  K  =  .Q16,  had  been  used  and  same  flux 
density  and  frequency  maintained,  what  would  have  been  the 
loss  in  above  machine? 

p_.016X12,0001-6 
107 

=  .00539  watt  per  cc.'at  1  cycle  per.sec. 
Total  loss   =  .00539  X  60  X  2500 
=  .81  K.W. 

Since  2500  cu.cms.  of  iron  weigh  less  than  45  Ibs.,  it  can  be 
seen  that  the  machine  is  not  very  large.  A  loss  of  .8  K.W.  then, 
would  be  fatal  to  it. 

For  methods  of  obtaining  the  hysteresis  loop,  see  the  Labor- 
atory Notes  on  the  experiment  with  this  title. 

Problem  29-6.  The  armature  of  a  motor  contains  70  Ibs.  of 
sheet  iron  (specific  gravity  7.7).  The  magnetic  flux,  which  has  a 
maximum  value  of  11,000  gausses,  goes  through  40  cycles  per.sec. 
What  is  the  K.W.  loss? 


154 


ELEMENTS  OF  ELECTRICITY 


SUMMARY  OF  CHAPTER  VI 


FIELD  INTENSITY  WITHIN  A  COIL;  AIR  CORE. 

TT_47rNI_i.26NI 
=~         = 


MAGNETOMOTIVE  FORCE. 


47TNI  t 

M= =  i.26NI. 

10 


Magneto- 
up  the 
motive 


These  two  equations  must  not  be  confused, 
motive  Force  is  the   magnetic   pressure   which 
magnetic    current.      It    'corresponds     to    the    El 
Force  which  sets  up  the  electric  current. 

RELUCTANCE.  The  reluctance  of  a  given  magnetic 
circuit  corresponds  to  the  resistance  of  an  electric  circuit. 

The  same  form  of  equation  is  used  to  find  the  reluctance 
of  a  certain  part  of  a  magnetic  circuit  as  is  used  to  find  the 
resistance  of  any  part  of  an  electric  circuit. 


Reluctances  in  series  are  added  like  resistances  in  series. 

Reluctances  in  parallel  are  treated  like  resistances  in  par- 
allel. 

OHM'S  LAW  OF  THE  MAGNETIC  CIRCUIT.  Ohm's 
Law  applies  to  the  magnetic  circuit  as  well  as  to  the  electric 
circuit. 


Current  (in  lines)  = 


Pressure 
Reluctance' 


(R 


VALUE  OF  IJL  DEPENDS  ON  B.  The  value  of  the  Per- 
meability /*  of  iron  and  steel  is  not  constant  like  the  value 
of  the  Conductivity  of  copper.  Every  change  in  the  Flux 


MAGNETIC  FIELD  DUE  TO  A   CURRENT          155 

Density  B  changes  the  value  of  the  Permeability  /*.  It  is 
therefore  necessary  to  know  at  what  Flux  density  a  piece 
of  iron  is  to  be  used,  before  the  correct  value  of  /j.  can  be 
determined. 

THREE  STAGES  OF  MAGNETIZATION.  When  a  piece 
of  iron  or  steel  contains  few  or  no  magnetic  lines,  it  requires 
considerable  magnetizing  force  to  set  up  an  increase  of  flux 
density.  After  starting  the  building  up  of  the  lines  within 
the  iron,  it  becomes  easy  to  increase  the  flux  until  a  point 
called  Saturation  is  reached.  While  it  is  possible  to  go  on 
increasing  the  flux  density  beyond  the  Saturation  point,  it 
is  highly  impracticable,  because  with  each  succeeding  increase 
in  flux  density,  a  greater  and  greater  magnetizing  force  must 
be  used,  until  soon  a  very  great  increase  in  magnetizing 
force  gives  but  a  small  increase  in  flux  density. 

The  Saturation  point  is  located  at  the  "  knee  "  of  the  Magnet- 
ization cmA.  A  machine  should  be  run  with  the  flux  density 
of  the  j^Roetic  circuit  at  about  the  Saturation  Point. 

HYSflRtESIS.  When  we  remove  the  magnetizing  force, 
all  the  magnetic  lines  do  not  fall  out,  because  there  is  a  certain 
"  lag  "  in  the  magnetic  flux  with  regard  to  the  magnetizing 
force.  This  "  lagging  "  is  called  "  Hysteresis." 

In  all  armature  cores  and  in  A.C.  machines  it  is  necessary 
to  expend  a  certain  amount  of  power  to  overcome  the  effects 
of  this  "  lagging."  Since  soft  iron  and  annealed  silicon- 
steel  exhibit  this  property  of  hysteresis  to  a  very  limited  degree, 
it  is  customary  to  use  these  materials  wherever  frequent 
and  large  changes  in  the  magnetic  flux  are  required. 

Steinmetz  gives  the  following  formula  for  computing  the 
"  Hysteresis  Loss  "  in  Watts  for  i  cu.cm.  of  iron  for  i  cycle 
per  second. 


p=- 


10' 


It  can  also  be  computed  from  the  area  of  the  Hysteresis 
Loop. 


156  ELEMENTS  OF  ELECTRICITY 


PROBLEMS    ON    CHAPTER    VI 

30-6.  A  wrought-iron  ring  200  cms.  long  is  to  carry  a  flux  of 
800,000  lines.  Flux  density  is  to  equal  6000  gausses,  (a)  What 
is  the  cross-section  area?  (b)  What  is  the  permeability  at  this 
density?  (c)  How  many  ampere-turns  are  required?  (d)  What 
is  reluctance  of  circuit? 

31-6.  In  order  to  magnetize  an  iron  rod,  a  magneto-motive 
force  of  400  units  is  necessary.  How  many  volts  must  be  applied 
to  a  coil  of  200  turns  and  68  ohms  resistance? 

32-6.  If  wire  in  Problem  31  is  No.  24,  B.  &  S.  copper,  how  many 
feet  of  it  are  in  the  coil? 

33-6.  It  is  desired  to  magnetize  a  40  cm.  cast  steel  ring  to  a 
density  of  16,500  lines  per  sq.cm.  800  turns  of  No.  23,  B.  &  S. 
copper  wire  are  used.  Average  length  of  each  turn  i^M  in.  How 
many  volts  must  be  applied  to  coil  in  order  to  set  upB^red  flux 
density? 

34-6.  How  many  ampere-turns  are  necessary  to  magnetize 
a  12-cm.  cast-steel  bar  of  6  sq.cms.  cross-section  so  that  it  has  a 
strength  of  2000  unit  poles?  Bar  is  bent  into  a  ring. 

35-6.  A  magnetic  circuit  is  made  up  of  300  cms.  of  wrought 
iron  50  sq.cm.  in  cross-section;  1.5  cms.  of  air,  55  sq.cms.  in 
cross-section.  F^nd  number  of  ampere-turns  necessary  to  set 
up  600,000  lines  of  force  in  above  circuit. 

36-6.  A  magnetic  circuit  is  made  up  of 

200  cms.  of  cast  steel,  40  sq.cms.  cross-section 
120    "      "  sheet   "      50    " 
1.2     "      "  air,  50     " 

Find  current  that  must  be  sent  through  a  coil  of  20,000  turns  in 
order  to  set  up  800,000  lines  in  above  circuit. 

37-6.  If  cross-section  of  each  part  of  circuit  in  Problem  36, 
were  doubled,  what  current  would  be  necessary? 

38-6.  A  magnetic  circuit  is  made  up  of : 

150  cms.  of  wrought  iron,  5  =  10,000  gausses. 
200    "      "sheet  steel,       5  =  14,000       " 
50     "      "  cast  iron,  B=   7,500       " 

1     "      "air,  #  =  12,000       " 

Find  ampere-turns  necessary  to  maintain  the  circuit  at  these  flux 
densities. 


MAGNETIC  FIELD  DUE   TO   A   CURRENT 


157 


39-6.  If  the  total  flux  in  circuit  of  Problem  38  is  1,000,000, 
what  must  be  the  area  of  each  part  of  the  circuit? 

40-6.   What  is  the  reluctance  of  circuit  in  Problem  39-6? 

41-6.  A  cast-iron  ring  has  a  mean  diameter  of  50  cms.  and  a 
cross-section  area  of  36  sq.cms.  How  many  ampere-turns  are 
required  to  set  up  a  flux  of  460,000  lines? 

42-6.  What  is  the  reluctance  and  permeability  of  ring  in  Problem 
41? 

43-6.  What  would  be  the  reluctance  and  permeability  of  ring 
in  Problem  41-6,  if  230,000  lines  were  set  up? 

44-6.  Fig.  129  represents  a  magnet  used  to  hoist  steel  rails. 
Magnet  core  A  is  of  wrought  iron,  120  sq.cms.  cross-section. 
Rail  B  has  a  cross-section  of  250  sq.cms.  There  is  a  space  of 
0.15  cm.  between  ends  of  core  A  and  rail  B,  due  to  rust.  The 
flux  required  is  1,800,000  lines.  How  many  ampere-turns  are 
required? 


— SQ-ems. f 


FIG.  129. 


FIG.  130. 


46-6.  The  lifting  magnet  Fig.  130  has  the  following  dimensions: 
Core  A,  mean  length  of  11  inches,  area  of  section  1  sq.in.  Core 
B,  mean  length  6.5  inches,  area  of  section  .9  sq.in.  Material 
of  both  is  sheet  steel.  Air  space  between  ends  of  A  and  B  £  in. 
If  100,000  lines  are  required  in  circuit  to  lift  B,  how  many  ampere- 
turns  must  be  wound  on  (7? 

46-6.  The  core  of  a  transformer  weighs  280  Ibs.  If  it  is  used 
on  an  alternating  current  circuit  of  133  cycles  per  sec.,  what  will 
the  hysteresis  loss  be?  #max=  10,000  gausses. 

47-6.  What  would  the  hysteresis  loss  be  in  Problem  46-6,  if 
the  transformer  were  used  on  a  60-cycle  circuit? 

48-6.  In  Fig.  156,  the  field  cores  are  of  wrought  iron  each 
50  cms.  long  and  an  average  cross-section  area  of  140  sq.cms. 
Yoke  is  of  cast  steel  25  cms.  long,  80  sq.cms.  cross-section;  armature 


158  ELEMENTS  OF  ELECTRICITY 

core  is  of  sheet  steel  14  cms.  long,  average  cross-section  of  120 
sq.cms.  Two  air  gaps  each  .8  cm.  long,  of  250  sq.cms.  cross-sec- 
tion area.  If  -1,250,000  lines  are  desired  in  air  gap,  of  how  many 
ampere-turns  must  the  field  coils  consist? 

49-6.  The  coils  in  Problem  48  each  have  8300  turns  of  wire. 
Each  lamp  in  a  circuit  supplied  by  this  generator  takes  .4  ampere. 
How  many  amperes  must  armature  deliver? 

50-6.  Armature  of  generator  in  Problem  48  makes  1200  R.P.M. 
What  is  hysteresis  loss? 

51-6.  Assume  1  sq.in.  of  curved  surface  of  coil  is  needed  to 
radiate  1  watt  to  prevent  a  coil  from  becoming  overheated. 
What  must  be  outside  diameter  of  coil  in  Problem  33-6  if  it  is 
to  be  used  continuously?  Length =25  cms. 

52-6.  On  basis  stated  in  Problem  51,  what  must  diameter  of 
coil  in  Problem  36  be  if  length  is  78  cms.?  Resistance=40  ohms. 

53-6.  On  basis  stated  in  Problem  51,  the  temperature  of  a  coil 
rises  55°  C.  above  surrounding  temperature  on  continuous  run. 
If  the  curved  surface  of  Problem  52  is  2750  sq.cms.,  how  high 
will  temperature  rise? 


CHAPTER  VII 
THE  GENERATOR 

Electromagnetic  Induction — Direction  of  Induced  E.M.F. — Amount 
of  Induced  E.M.F. — Current  in  Revolving  Loop — Sine  Curve  of 
E.M.F. — Collecting  Rings — A. C.  Power — Commutator — D.C.  Power 
— Ring  and  Drum  Armatures — Action  within  an  Armature 
— Magnetization  of  the  Core — Neutral  Axis — Axis  of  Least 
Sparking — Voltage  and  Resistance  of  D.C.  Armature — Field 
Excitation  of  Generator — Separately  Excited,  Series  Wound, 
Shunt  Wound,  Compound — Losses  in  Generator— Eddy  Current 
Loss. 

THUS  far  we  have  been  studying  the  laws  controlling 
the  action  of  electric  and  magnetic  pressures  and  currents. 
We  are  now  to  consider  how  electric  pressures  and  currents 
are  generated  and  maintained. 

104.  Electromagnetic  Induction.  It  has  been  shown 
that  we  always  set  up  a  magnetic  field  around  a  con- 
ductor whenever  we  set  up  an  electric  current  in  the  con- 
ductor. This  magnetic  field  lasts  as  long  as  the  current  lasts. 

On  the  other  hand,  it  has  been  found  that  if  we  have  a 
magnetic  field  present,  and  wish  to  set  up  an  electric  current 
in  a  conductor,  it  is  necessary  to  move  the  conductor  in  such 
a  way  as  to  cut  the  magnetic  lines.  For  instance,  if  a  wire 
is  swung  in  the  earth's  magnetic  field  in  such  a  way  as  to 
cut  across  the  lines,  we  find  that  an  E.M.F.  is  induced  in 
the  wire.  This  is  shown  by  placing  a  galvanometer  across 
the  ends  of  the  wire  and  noting  the  deflection  every  time 
the  wire  is  swung.  As  long  as  the  wire  remains  stationary 
no  current  flows.  In  fact,  even  if  the  wire  moves,  but  in 

159 


160 


ELEMENTS  OF  ELECTRICITY 


a  direction  parallel  to  the  lines  of  force,  no  current  flows. 
This  shows  that  the  conductor  must  move  and  cut  lines  of 
force  in  order  to  have  an  E.M.F.  induced  in  it. 

105.  Direction  of  Induced  E.M.F.  Consider  Fig.  131.  If 
the  wire  AB  is  moved  down  across  the  lines  of  force  as 
marked,  an  electric  current  will  flow  along  the  wire  from  A 
to  B.  If  the  wire  were  moved  up  across  the  lines,  the 
current  would  flow  in  the  opposite  direction,  from  B  to  A. 


FIG.  131. — Direction  of  E.M.F.  induced  in  wire  cutting  magnetic  lines  of  force. 

The  rule  for  finding  the  direction  of  the  Induced  E.M.F. 
which  sets  up  this  current,  is  as  follows. 

Extend  the  THUMB,  FOREFINGER  and  MIDDLE  FINGER 
of  the  RIGHT  hand  at  right  angles  to  one  another.  Let 
the  THUMB  point  in  the  direction  of  the  motion,  the  FORE- 
FINGER in  the  direction  of  the  magnetic  flux,  then  the 
MIDDLE  FINGER  will  be  pointing  in  the  direction  of  the 
induced  E.M.F. 

The  hand  in  Fig.  131  shows  the  application  of  this  rule 
to  the  case  of  the  wire  being  moved  down. 

Fig.   132  brings  out  this  relation  more  clearly. 

The  direction    of    the  induced  E.M.F.  then,  is  seen  to 


THE  GENERATOR  161 

depend  upon  the  direction  of  the  flux  and  the  direction  of 
the  motion  of  the  conductor. 

106.  Amount  of  Induced  E.M.F.      The    amount  of  the 
induced   E.M.F.    depends   upon   the  number   of   lines   cut 
per  second.     It  has  been  found  that 
if    a    conductor    cuts    100,000,000 
lines  per  sec.,  that  an  E.M.F.  of  1 
volt  is  produced  between  the  termi- 
nals.    This  number  100,000,000  is 
usually  written   108,   and   we  read 
it  "  ten-to-the-eighth-power." 

Example  21.  A  wire  passes  40  times 
a  second  across  the  pole  face  of  a  field 
magnet  the  flux  density  of  which  is 

15,000    gausses.      The  dimensions  of  the  face  are   30X20  cms. 
What  E.M.F.  is  induced  in  the  wire? 

(j>  =  600  X  15,000  =  9,000,000 = 9  X 106. 
Lines  cut  per  sec.  =  9Xl06X40  =  3.6X108. 

3.6  X108 
Volts  =  — — g — -3.6  volts. 

Problem  1-7.  How  many  volts  are  induced  across  the  ends  of 
a  wire  which  cuts  3X1010  lines  in  2.4  sees.? 

Problem  2-7.  A  wire  cuts  through  a  field  of  2,000,000  lines,  at 
an  average  rate  of  2400  times  per  minute.  How  many  volts 
(average)  are  set  up  across  the  wire? 

Problem  3-7.  A  wire  100  cms.  long  passes  through  a  magnetic 
field  where  the  flux  density  is  8000  gausses.  If  the  velocity  at 
right  angles  to  the  lines  is  1200  cms.  per  sec.  what  voltage  is  in- 
duced in  the  wire? 

Problem  4-7.  If  a  wire  which  is  cutting  through  a  magnetic  field 
at  the  rate  of  100  ft.  per  sec.  induces  4  volts -across  its  terminals, 
how  fast  must  it  move  to  set  up  6  volts? 

107.  Current  in  Revolving  Loop.  The  E.M.F.  of  a 
generator  is  the  E.M.F.  induced  in  wires  revolving  so  as 
to  cut  the  lines  of  force  of  a  powerful  magnetic  field.  Let 
us  consider  the  case  of  a  single  loop  of  wire  so  revolved. 


162 


ELEMENTS  OF  ELECTRICITY 


In  Figs.  133  and  134,  the  loop  of  wire  is  revolving  as  shown 
by  the  arrow. 

In  Fig.   133,  the  side  of  the  loop  BD  is  moving  down 

across  the  lines.  By  apply- 
ing the  Right- Hand  Rule,  the 
induced  E.M.F.  is  found  to  be 
in  the  direction  of  D  to  B 
as  marked.  The  side  AC  is 
moving  up  and  the  induced 
E.M.F.  is  from  A  to  C  as 

FIG.  133. — Model  single  loop  generator. 

marked.     Thus  a  current  will 

flow  around  the  loop  in  the  direction  ACDB,  as  marked. 
In  Fig.  134,  the  wire  BD  has  now  turned  to  the  opposite 
side,  and  is  moving  up  instead  of  down.  The  induced 
E.M.F.  has  now  reversed  and  is  from  B  to  D.  For  the 
same  reason,  the  E.M.F.  in  AC  is  reversed  and  is  from  C 
to  A.  This  now  causes  a  current  to  flow  around  the  coil 
in  the  direction  BDCA,  which  is  the  reverse  of  the  direc- 
tion of  the  current  when  the  coil  was  in  the  other  position. 


FIG.  134. — E.M.F.  in  loop,  after  one-half  turn. 

Thus  the  current  in  a  revolving  coil  flows  one  way  dur- 
ing half  the  revolution  and  in  the  reverse  direction  during 
the  other  half. 

There  is  then  an  alternating  current  in  a  closed  loop  of 
wire  revolved  in  a  magnetic  field. 

The  current  will  be  at  its  maximum  value  in  one  direc- 
tion when  the  coil  is  in  the  position  of  Fig.  133,  and  at  its 
maximum  value  in  the  opposite  direction,  when  in  the  posi- 
tion of  Fig.  134,  because  at  these  positions  the  sides  of  the 
coils  are  cutting  the  greatest  number  of  lines  per  sec.  When 


THE  GENERATOR 


163 


the  loop  is  vertical,  as  in  Fig.  135,  the  sides  of  the  loop  are 
moving  parallel  to  the  magnetic  lines  and  are  not  cutting 
them.  There  is  then  no  E.M.F.  induced  in  the  loop.  It  is 
then  said  to  be  in  the  "  neutral  position." 

108.  Sine  Curve  of  E.M.F.     If  we   consider  the  neutral 
position  of    the  coil   (Fig.   135),   as  the  "  zero  "  position, 


FIG.  135. — Loop  in  neutral  position.    The  sides  are  not  cutting  lineg. 

then  the  horizontal  position  of  Fig.  133  would  be  90°  from 
the  zero  position.  When  the  coil  had  become  vertical 
again,  it  would  be  again  in  a  neutral  position,  at  180°  from 
the  first  neutral  position.  When  again  horizontal,  as 
in  Fig.  134,  it  would  be  270°  from  the  first  neutral  position. 
We  may  plot  a  curve  using  the  position  (in  degrees  from  the 
neutral  position)  as  the  abscissae  and  the  induced  E.M.F.  as 
the  ordinates. 


4-30 
o+lO 

a    o 

t 

S* 

X 

^ 

^ 

. 

/ 

\ 

s 

y 

/ 

/ 

f\ 

\ 

0 

(>° 

I 

1 

1 

si-« 

-30 

v 

\ 

/ 

x 

^ 

POSITION  OF  LOOP   IN  DEGREES 

FIG.  136.  —  Sine  curve  of  induced  E.M.F. 


When  the  E.M.F.  is  in  one  direction,  we  call  it  +,  and  in 
the  other,  when  the  field  is  uniform—. 

We  obtain  a  curve  as  in  Fig.  136. 

This  curve  shows  that  the  voltage  (induced  E.M.F.) 
rises  rapidly  to  a  maximum  at  the  90°  position  (horizontal) 


164  ELEMENTS  OF  ELECTRICITY 

where  the  lines  are  being  cut  at  the  greatest  rate.  It 
falls  again  to  zero  at  the  180°  position  (vertical)  when  the 
sides  of  the  loop  are  moving  parallel  to  the  lines  and  are 
thus  not  cutting  them.  Then  the  sides  begin  cutting  in 
the  opposite  direction  and  a  voltage  is  thus  induced  in  the 
opposite  direction  (which  we  have  agreed  to  call  "  negative.") 
The  voltage  in  this  direction  increases  rapidly  to  a  maximum 
at  the  270°  position  (horizontal)  where  the  lines  are  again 
being  cut  at  the  greatest  rate  (only  in  the  opposite  direc- 
tion) .  Again  it  decreases  as  it  begins  to  approach  the  neu- 
tral (vertical)  position  where  the  sides  again  move  parallel 
to  the  lines,  and  thus  again  at  this  point  no  voltage  is  in- 
duced. This  "  cycle  "  of  events,  as  it  is  called,  takes  place 
every  complete  revolution  in  a  two-pole  machine,  the 
maximum  voltage  reached  in  'each  direction  depending 
upon  the  number  of  lines  cut  per  sec.,  at  the  instant  the 
sides  are  cutting  at  the  greatest  rate. 

In  order  to  get  a  high  maximum  voltage  in  a  generator: 

First.  The  inside  of  the  loop  is  filled  with  an  iron  core 
to  increase  and  concentrate  the  magnetic  field. 

Second.  A  great  many  loops  of  wire  are  wound  in 
series  on  this  core,  so  that  the  voltage  across  the  ends 
at  any  time  is  the  sum  of  the  voltages  in  each  loop. 

The  "  loops  "  and  the  "  core,"  together  with  the  device 
for  taking  out  current  from  the  loops,  form  the  armature 
of  most  Direct  Current  and  many  Alternating  Current 
Generators. 

The  current  within  the  armature  of  a  DIRECT  as  well  as 
of  an  ALTERNATING  CURRENT  GENERATOR  is  always  an 
ALTERNATING  Current,  and  goes  through  much  the  same 
"  cycle  "  of  values  described  by  the  curve  in  Fig.  136.* 

Whether  a  machine  delivers  an  Alternating  or  a  Direct 
Current,  then,  must  depend  upon  the  device  used  for  taking 

*  The  one  exception  to  this  statement  is  the  UNIPOLAR  generator, 
which  has  come  into  some  use  with  the  advent  of  the  high-speed  steam 
turbine. 


THE  GENERATOR 


165 


Fm.  137.— Single  loop  fitted  with  col- 
lecting rings.  An  alternating  cur- 
rent is  delivered. 


off  the  current  from  the  armature.  This  device  may  consist 
of  either  COLLECTING  RINGS,  which  deliver  an  alternating 
current,  or  a  COMMUTATOR,  which  delivers  a  direct  current. 

109.  Collecting    Rings.     If  we  have  the  outside  circuit 
continually   connected  with  the  two  ends  of  the  revolving 
coils,  we  must   get  the  same 

kind  of  voltage  induced  across 

the  outside  circuit  that  we  do 

across    the    inside.      That   is, 

we  would   get  an  alternating 

current    from    the    machine, 

since    the    inside    current    is 

always    alternating.      This    is 

done  by  connecting  the  ends 

of  the  wire  of  the  coils  to  rings,  called  COLLECTING  RINGS. 

Brushes  bearing,  on  these  rings  will  lead  the  alternating 

current  to  any  desired  outside  circuit. 

Fig.  137  shows  the  principle  of  Collecting  Rings.  Ci,  C2 
represent  the  sides  of  a  coil  of  wire 
in  the  armature;  RI  and  R2  the 
collecting  rings  joined  one  to  each 
end  of  the  coil;  BI  and  B2,  the 
brushes,  and  L,  a  bank  of  lamps  as 
outside  circuit. 

Since  the  lamps  always  form  a 
continuous  circuit  with  the  coil, 
they  will  always  have  the  same 
E.M.F.  that  the  coil  has.  The  sine 
curve  of  Fig.  136  will  then  be  a  fair 
representation  of  the  E.M.F.  across 
the  lamps  as  well  as  across  the  arma- 
ture. A  single-coil  MAGNETO  is  an 
example  of  a  very  simple  Alternator, 
and  is  shown  in  Fig.  137a. 

110.  Commutator.     If  we  can  connect  each  brush  first 
with  one  end  and  then  with  the  other  end  of  the  revolving 


FIG.  137  a. — Magneto 
generator. 


166  ELEMENTS  OF  ELECTRICITY 

coil,  making  the  change  just  at  the  instant  the  current  in 
each  side  is  reversing,  we  shall  then  have  the  current  on 
the  outside  always  in  the  same  direction.  This  is  accom- 
plished by  means  of  a  COMMUTATOR.  Fig.  138  shows  the 
principle  of  the  commutator.  The  segments  S\  and  $2 
are  attached  each  to  an  end  of  the  revolving  coil  C\,  C2. 
The  brushes  B±  and  B2  collect  the  current  and  deliver  it  to 
the  lamp  bank  L. 

Assume  that,  at  the  instant,  the  current  in  Ci  is  coming 
to  the  brush  B\.  The  current  in  the  side  C2  must  be  going 
in  the  opposite  direction  or  away  from 
the  brush  B2.  BI  is  then  +  and  B2 
is  — .  The  brush  BI  bears  on  the 
segment  Si  as  long  as  the  current  in 
Ci  is  in  this  direction.  As  soon  as  the 
current  in  Ci  changes  to  the  oppo- 
site direction,  the  segment  S\  has  left 
the  brush  BI  and  the  segment  S2  has 
come  into  contact  with-B^  But  just 
as  the  current  in  C\  has  reversed,  so 

FIG.  138.— Single  loop  fit-  .  x 

ted  with  commutator.  A     the  current  in  C2  has  also   reversed, 

direct  current  is  delivered. 

and  the  current  now  flows  to  segment 
S2.  S2  now  delivers  current  to  brush  BI  which  keeps  it 
still  the  positive  brush.  Thus  BI  is  always  positive.  In 
the  same  way  it  is  seen  that  B2  will  always  be  the  negative 
brush.  The  current  delivered  to  the  lamps  then  will  always 
be  in  the  same  direction. 

A  single  loop  of  wire,  of  course,  does  not  deliver  a  steady 
current,  even  with  the  commutator,  although  the  current 
is  always  in  the  same  direction.  The  brushes  are  so  placed 
that  the  change  from  one  segment  to  another  is  made  when 
there  is  no  current  flowing  in  the  coil.  At  this  moment, 
also,  there  would  be  no  current  flowing  in  the  lamps.  So 
we  would  have  a  Pulsating  Current  in  the  lamps  as  repre- 
sented in  Fig.  139.  The  current  on  the  inside  of  the  coil 
is  still  that  represented  by  the  curve  in  Fig,  136.  The  values 


THE  GENERATOR 


167 


of  the  voltages  on  curve  in  Fig.  139  are  the  same  as  those 
on  curve  in  Fig.  136,  the  voltage  across  the  lamps  being 
zero  when  the  voltage  across  the  coil  is  zero,  and  greatest 
when  greatest  across  the  coil.  The  sign,  however,  for  the 
voltage  across  the  lamps  is  always  positive  (+),  since  the 
current  is  always  in  the  same  direction.  Curve  139  then 
is  Curve  136,  with  the  negative  loops  turned  into  positive 
ones. 

To  obtain  a  steady  direct  current,  a  large  number  of  coils 
is  placed  around  the  core,  and  the  commutator  divided 
up  into  a  correspondingly  large  number  of  segments.  The 
loops  of  wire  of  the  different  coils  are  distributed  so  that  at 


+30 
2+20 

i?  +io 

.2       o 

5-20 
-30 

X" 

^ 

^ 

X 

^ 

^ 

i 

S 

i 

/ 

\ 

\ 

^ 

/ 

/ 

\ 

/ 

« 

6- 

4« 

I 

J° 

POSITION    OF   LOOP    IN    DEGREES 

FIG.  139. — Curve  of  pulsating  E.M.F.  across  single  coil  armature  fitted  with 
commutator. 

every  instant  some  of  them  are  cutting  lines  at  the  max- 
imum rate.  Examples  of  the  more  important  of  these 
arrangements  will  be  given  later. 

It  can  be  stated  as  a  general  rule  that  an  armature  always 
maintains  an  alternating  pressure  across  its  coils.  When 
used  with  COLLECTING  RINGS,  the  brushes  are  then  placed 
always  across  the  two  ends  of  the  same  coil  and  thus  the 
machine  delivers  an  alternating  current.  When  used  with 
a  commutator,  each  brush  is  placed  so  that  it  takes  the 
current  from  one  side  of  the  coil  only  so  long  as  that  side 
is  in  a  certain  part  of  the  field.  When  the  current  reverses 
in  one  side  of  the  coil,  that  side  is  passed  on  to  the  other 
brush.  In  this  way  a  current  is  delivered  which  is  always 


168 


ELEMENTS  OF  ELECTRICITY 


in  the  same  direction.  Machines  using  or  delivering  such 
a  current  are  called  Direct  Current  Machines.  It  is  this 
type  that  we  will  now  study  more  in  detail. 


Fig.  140  shows  a  mounted  armature  fitted  so  that  it  can 
deliver  both  direct  and  alternating  current.  On  the  left-hand 
side  of  the  machine  are  seen  the  collecting  rings.  Brushes 


THE  GENERATOR  169 

placed  on  these  rings  will  have  an  alternating  E.M.F. 
across  them  when  the  armature  is  revolved  in  a  strong 
field  as  explained  above.  On  the  right  is  a  commutator 
with  a  large  number  of  segments.  Brushes  placed  on  the 
commutator  at  proper  positions  will  have  a  direct  E.M.F. 
across  them,  and  since  there  is  a  large  number  of  coils  in 
the  armature,  the  E.M.F.  will  be  very  steady. 

Fig.  141  shows  a  partly-wound  Direct  Current  armature. 


FIG.  141. — Partly-wound  armature  of  D.C.  generator.     Allis-Chalmers. 


111.  Ring  and  Drum  Armature.  In  making  an  armature, 
there  are  two  general  types  of  cores  on  which  the  loops  of 
wire  are  wound,  (1)  the  DRUM  and  (2)  the  RING.  The 
DRUM  type  consists  of  a  cylinder  of  iron  built  up  of  thin 
sheets  cut  to  proper  shape  in  a  punch  press. 

The  wires  are  then  wound  around  the  outside  of  this 
drum,  as  in  Fig.  142,  which  represents  a  four-loop  armature. 


FIG.  142. — Drum-wound  armature. 


The  ring  core  is  similar  to  the  drum,  except  that  it  is 
hollow  and  the  wires  are  wound  in  and  out  and  do  not  pass 


170 


ELEMENTS  OF  ELECTRICITY 


all  around  the  outside.  Fig.  143  shows  the  ring  method 
of  constructing  an  armature. 

Note  that  in  a  DRUM-wound  armature  both  sides  of  each 
loop  cut  the  lines  of  force,  while  in  the  Rixo-wound  one  side 
only  cuts  lines,  the  other  side  being  on  the  inside  of  the  ring, 
which  is  a  space  practically  free  from  magnetic  lines.  See 
Fig.  145  for  illustration  of  this  fact. 

Most  cores  are  now  made  of  the  drum  type,  because  it 
is  simple  in  construction  and  mechankally  stronger.  The 
surface  of  such  a  core  is  generally  not  smooth,  but  slotted 
so  that  the  loops  of  wire  lie  in  grooves  between  projections 


FIG.  143. — Ring-wound  armature.  FIG.  144. — Slotted  armature  core, 

drum  type. 

called  teeth.  Fig.  144  shows  an  ideal  cross-section  of  such 
a  slotted  armature. 

Although  the  ring  type  is  very  rarely  used,  it  affords  the 
simplest  diagram  from  which  to  explain  the  action  in  an 
armature.  The  action  in  a  drum  armature  is  essentially 
the  same,  but  it  is  more  difficult  to  represent  it  without 
complicated  diagrams. 

112.  Actions  within  an  Armature.  Magnetization  of 
the  Core.  When  a  soft  iron  ring  is  placed  in  the 
magnetic  field  between  the  poles  of  a  2-pole  machine,  the 
magnetic  field  may  be  represented  as  in  Fig.  145. 

Suppose  now  this  soft  iron  ring  were  wound  with  six 
coils  of  wire  and  connected  to  six  commutator  segments 
as  in  Fig.  146.  If  this  armature  were  rotated  in  the  direc- . 


THE  GENERATOR 


171 


tion  shown,  the  outside  wires  would  cut  lines  of  force, 
and  there  would  be  a  tendency  for  an  electric  current  to 
flow  through  the  coils  in  the  directions  marked.  Test 
this  by  the  Right-Hand  Rule. 


FIG.  145. — Field  in  core  of  ring  arma- 
ture when  there  is  no  current  in  the 
armature  coils. 


FIG.  146. — Dash  lines  show  direction 
of  field  set  up  by  current  in  arma- 
ture coils. 


A  magnetic  field  would  then  be  set  up  within  the  iron 
ring  in  the  direction  of  the  dash  lines,  i.e.,  the  bottom  of 
the  ring  would  tend  to  become  a  north  pole.  We  now 


Field  of  Poles 


FIG.  147. — Resultant  or  distorted  field  in  armature  core. 

would  have  two  fields  at  right  angles  to  each  other.     The 
resultant  field  is  then  the  combination  of  the  two,  shown 


172  ELEMENTS,  OF  ELECTRICITY 

in  Fig.  147,  just  as  in  mechanics  a  single  force  is  often  the 
resultant  of  two  forces  at  right  angles  to  each  other. 

When  any  current  flows  in  the  armature  and  sets  up  a 
cross  field  in  the  armature  core,  the  field  is  then  said  to  be 
distorted. 

The  angle  I  measures  the  amount  the  field  is  distorted. 
This  anglo  lies  between  ab,  a  line  drawn  perpendicular  to 
the  original  field,  and  bisecting  angle  between  axes  of  poles, 
and  cd,  a  line  drawn  perpendicular  to  the  resultant  field. 

113.  Setting  the  Brushes.  Neutral  Axis.  Axis  of  Least 
Sparking.  The  line  ab,  Fig.  147,  is  called,  the  Neutral  Axis, 
because  it  is  perpendicular  to  the  field  at  no  load.  It 
indicates  the  places  in  which  a  wire  would  be  cutting 
no  lines  of  force,  if  the  armature  were  revolving  but  de- 
livering no  current.  The  line  cd  indicates  the  approximate 
positions  of  the  brushes,  in  order  that  there  be  no  sparking 
when  the  armature  is  delivering  a  current.  This  is  called 
the  Axis  of  Sparkless  Commutation  and  is  best  explained 
by  referring  to  Fig.  148. 

Note  the  arrangement  of  the  six  coils  in  Fig.  148.  Each 
is  connected  across  the  gap  in  the  commutator,  the  ends 
being  joined  to  adjacent  segments.  The  segments  then 
serve  to  connect  all  the  coils  in  series. 

The  brushes  BI  and  B2  are  placed  180°  apart  on  about 
the  axis  CD,  perpendicular  to  the  resultant  field.  Notice 
that  in  a  generator,  this  axis  cd  is  ahead  of  the  neutral  axis 
ab  which  is  perpendicular  to  line  joining  poles.  This  axis 
of  commutation  CD  will  be  seen  to  be  behind  the  neutral  axis 
ab  in  a  motor.  The  words  "  ahead  "  and  "  behind,"  refer 
to  the  position  relative  to  the  motion  of  the  armature.  If 
the  motion  were  in  the  opposite  direction,  then  the  line  cd 
would  be  said  to  be  behind  the  line  ab. 

By  applying  the  Right-Hand  Rule  to  the  outside  wires 
in  coils  2  and  3  (the  inside  wires  cut  no  lines)  we  find  that 
the  current  tends  to  flow  from  right  to  left  across  the  face 
of  the  ring,  coming  out  of  the  face  of  the  ring  in  such  a  way 


THE  GENERATOR 


173 


as  to  flow  away  from  brush  BI,  and  toward  brush  B2.  The 
current  in  coils  5  and  6  goes  into  the  face  of  the  ring,  but 
away  from  BI  and  toward  B2.  Thus  there  are  two  currents, 
one  in  each  half  of  the  armature  flowing  in  parallel  away 
from  Bl  and  toward  B2.  The  brush  B±  would  then  be 
negative  and  B2  positive,  since  B2  would  deliver  current 
to  the  outside  circuit,  and  BI  receive  it  back  to  the  arma- 
ture. 

The  brushes,  however,  form  a  short  circuit  across  the 
coils  1  and  4,  when  they  are  in  just  this  position.     But 


FIG.  148. — Position  of  brushes  in  D.C.  generator. 

note  that  in  just  this  position  these  coils  are  cutting  no  lines 
of  force  and  there  is,  therefore,  no  current  flowing  in  them. 
This  is  the  chief  reason  for  the  brushes  being  placed  at  this 
particular  place,  i.e.,  at  axis  of  sparkless  commutation. 
Suppose,  for  instance,  that  the  brushes  were  placed  on  the 
neutral  axis  ab.  When  the  coils  reached  this  position, 
they  would  be  cutting  lines  of  force  and  as  the  brushes 
would  momentarily  short-circuit  them,  a  current  would 
flow,  which  might  be  large  enough  to  damage  the  coil. 
The  greatest  harm,  however,  would  be  done  when  the 
shorted  coil  moved  out  from  under  the  brush  and  this  cur- 
rent in  it  were  broken.  The  breaking  of  the  current  would 


174  ELEMENTS  OF  ELECTRICITY 

cause  an  arc  between  the  brush  and  commutator  segment 
which  would  soon  roughen  and  destroy  the  commutator. 

It  would  appear  from  the  above  that  in  order  to  secure 
sparkless  commutation,  the  brushes  must  be  placed  in  such 
a  position  that  such  coils  only  are  short-circuited  as  are 
cutting  no  lines  of  force,  and  thus  when  the  short  circuit 
is  broken,  there  is  no  current  to  be  broken  and  thus  no  arc 
is  formed.  This  would  place  the  brushes  exactly  on  the 
axis  of  sparkless  commutation. 

There  is,  however,  another  point  to  be  considered  which 
causes  this  axis  to  be  slightly  off  from  a  line  perpendicular 
to  the  resulting  field. 

Consider  brush  B2.  Coil  3,  which  has  not  yet  reached 
the  brush,  is  delivering  current  to  it.  Coil  5,  which  has 
just  passed  away  from  the  brush,  is  also  delivering  current 
to  it.  Yet  the  current  in  Coil  5  is  flowing  in  the  opposite 
direction  to  the  current  in  Coil  3.  The  current  in  3  is  com- 
ing out  of  the  face  of  the  ring,  the  current  in  5  is  going  into 
the  face  of  the  ring.  Thus  the  current  in  a  coil  must  not 
only  stop  flowing  when  it  reaches  the  brush,  but  also 
must  reverse  in  direction  on  leaving  it.  In  order  to  stop 
the  current  flowing  ii  this  coil  it  has  been  found  advisable 
to  set  up  an  E.M.F.  in  the  opposite  direction.  See  Chapter 
X  on  INDUCTION.  This  can  be  done  easily  if  the  brush 
is  set  a  little  ahead  of  the  neutral  axis,  so  that  the  coil  has 
begun  to  cut  lines  in  the  opposite  direction  before  it  is 
short-circuited.  The  E.M.F.  thus  induced  is  in  the  opposite 
direction  to  the  current  that  is  flowing  and  thus  quickly 
reduces  that  current  to  zero.  Before  this  E.M.F.  can  set 
up  a  current  of  its  own  in  the  opposite  direction  in  the  coil, 
the  brush  has  ceased  to  short-circuit  it,  and  thus  there  is 
no  "  short-circuit  current  "  to  be  broken.  In  this  position 
the  brush  is  said  to  lie  on  the  "Axis  OF  COMMUTATION," 
which  in  a  generator  is  usually  a  little  ahead  of  the  line 
CD.  The  angle  between  the  NEUTRAL  Axis  and  the  Axis 
OF  COMMUTATION  is  called  the  "ANGLE  OF  LEAD." 


THE   GENERATOR 


175 


For  certain  reasons  concerning  the  magnetic  disturbances 
which  this  Lead  of  the  brushes  causes,  it  should  always 
be  made  as  small  as  possible. 

The  actions  taking  place  inside  of  a  6-coil  armature  in  a 
2-pole  generator  have  been  described  because  they  are 
typical  of  all  D.C.  Generators,  whether  Drum  or  Ring 


FIG.  149. — Arrangement  of  poles  in  a  six  pole  Westinghouse  generator. 

wound,  bipolar  or  multipolar.  It  would  not  be  advisable 
to  go  into  the  different  details  of  armatures  of  other  types 
or  into  further  details  in  this  type,  since  these  belong  to 
a  more  advanced  course. 

114.  Multipolar    Generators.      In    order    to    obtain    a 
voltage  high  enough  for  practical  purposes,  the  armature 


176  ELEMENTS  OF  ELECTRICITY 

of  a  2-pole  generator  has  to  be  run  at  speeds  which  are 
prohibitive  for  machines  of  any  size.  Since  the  voltage 
depends  upon  the  number  of  lines  cut  per  sec.,  when  great 
speed  is  not  permissible,  more  lines  of  force  must  be  added 
to  the  field.  This  is  usually  done  by  adding  more  poles. 
Fig.  37  shows  a  4-pole  generator.!  Notice  that  when  the 
armature  makes  one  revolution,  each  conductor  cuts  the 


FIG.  150. — A  six  pole  generator  with  armature  in  place.    Westinghouse. 

lines  in  the  air  gap  four  times  instead  of  twice,  as  in  a  2-pole 
machine.  The  speed  can  therefore  be  slowed  down  in 
proportion  and  practically  the  same  voltage  obtained, 
other  conditions  remaining  unchanged.  Fig.  149  shows 
the  arrangement  of  the  poles  in  a  6-pole  geneiator. 

Fig.  150  shows  a  6-pole  generator  with  armatuie  in  place. 
Notice  that  every  other  brush  is  connected  to  the  same 
terminal,  making  three  positive  brushes  and  three  negative. 
Fig.  150a  shows  this  diagrammatically. 

115.  Resistance  of  a  D.C.  Armature.  It  has  been 
shown  that  in  a  2-pole  generator,  the  current  flows  in  two 


THE  GENERATOR 


177 


parallel  circuits  within  the  armature,  since  the  brushes 
put  the  two  halves  of  the  armature  in  parallel.  Thus  the 
resistance  of  the  armature  is  one-half  the  resistance  of  either 
of  these  two  circuits.  If  there  were  4  poles  to  the  machine 
and  4  brushes,  the  current  would  flow  in  4  parallel  circuits, 
and  the  ARMATURE  RESISTANCE  would  be  one-fourth  that 
of  any  single  path  or  circuit  through  the  armature. 


FIG.  150a. — Brush  connection  in  a  six  pole  generator. 

Example.  The  armature  of  a  4-pole  4-brush  D.C.  generator 
is  wound  with  360  ft.  of  No.  18  (B.  &  S.  gauge)  copper  wire.  What 
is  the  armature  resistance  at  68°  F.? 

Resistance  per  1000  ft.  of  No.  18  =  6.37  ohms, 
of      360  "    "    "      "  -2.29  ohms. 


"  1  circuit  -^— - 

.573 
armature  — : — 


.573  ohm. 
.143  ohm. 


Problem  5-7.  A  six-pole  6-brush  generator  has  800  ft.  of  No 
4  B.  &  S.,  copper  wire  wound  on  the  armature,  (a)  What  is  the 
total  resistance  of  the  wire  wound  on  the  armature  (20°  C.)? 

(6)  What  is  the  resistance  of  one  path  through  the  armature? 

(c)  What  is  the  armature  resistance  from  brush  to  brush? 


178  ELEMENTS  OF  ELECTRICITY 

Problem  6-7.  What  will  armature  resistance  of  generator  in 
Prob.  5  become  when  the  machine  has  been  running  a  while  and 
the  temperature  has  risen  50°  C.? 

Problem  7-7.  The  armature  resistance  of  a  bipolar  generator 
was  measured  at  20°  C.  and  found  to  be  2.3  ohms.  Size  wire  used 
was  No.  14  B.  &  S.  How  many  feet  of  wire  were  there  on  the 
armature? 

Problem  8-7.  What  is  armature  resistance  of  generator  Fig. 
150,  if  740  ft.  of  No.  6  B.  &  S.  copper  are  used  in  winding  armature? 

116.  Voltage  across  a  D.C.  Armature.  The  voltage 
induced  in  a  D.C.  armature  is  the  sum  of  the  voltages 
induced  at  a  given  instant  in  the  separate  coils  forming 
one  circuit  of  the  armature. 

In  Fig.  148,  the  voltage  across  the  brushes  at  the  instant 
would  be  the  voltage  across  coil  2  plus  the  voltage  across 
coil  3. 

Generally  all  the  coils  are  not  cutting  lines  at  the  same 
rate,  and  thus  the  voltage  across  each  coil  is  different. 
Each  coil  then  does  not  contribute  an  equal  amount  to  make 
up  the  voltage  across  the  brushes. 

The  average  voltage  across  the  brushes  depends  upon 
how  many  lines  are  cut  per  sec.  by  the  wires  forming  one 
circuit  of  the  armature  between  the  brushes. 

Therefore,  to  find  the  voltage  which  any  D.C.  armature  is 
generating,  it  is  merely  necessary  to  find  the  number  of 
lines  of  force  that  is  being  cut  per  sec.  by  those  conductors 
only  which  lie  between  two  adjacent  brushes,  since  these 
conductors  form  one  path  through  the  armature. 

Example.  Assume  that  there  are  198  active  conductors  on 
armature  of  Fig.  150.  Speed  is  800  R.P.M.  Each  pole  has 
5,000,000  lines.  What  E.M.F.  is  generated? 

The  number  of  conductors  between  any  two  adjacent  brushes, 
that  is,  in  one  path  in  armature,  must  be : 

108 

—  =  33  conductors. 


THE  GENERATOR  179 

Voltage  induced  in    these    conductors    equals  lines  cut  per  sec. 
divided  bv  108. 


E 


L 

_  33  X  5,000,000  X  6  X  800 

~~;      108X60 
=  132  volts. 


Problem  9-7.  Assuming  resistance  found  in  Prob.  8  as  correct 
for  armature  in  above  example,  what  will  be  the  brush  potential 
of  the  generator  when  delivering  280  amperes? 

Problem  10-7.  The  armature  of  a  4-pole  4-brush  generator 
consists  of  800  active  conductors.  Speed  =  1200  R.P.M.  <j)  for 
each  pole  is  6,000,000  lines.  Find  E.M.F. 

Problem  11-7.  Armature  in  Prob.  5-7  has  200  active  con- 
ductors. R.P.M.  =  740.  0  =  4,000,000  for  each  pole.  What  is  the 
E.M.F.? 

Problem  12-7.  What  is  the  terminal  voltage  of  generator  in 
Problem  11  when  delivering  400  amperes?  Temperature  is  50°  C. 

Problem  13-7.  What  is  the  terminal  voltage  of  a  bipolar 
generator,  the  drum  armature  of  which  has  960  active  conductors? 
Speed  =  1800  R.P.M.  <£  =  8,000,000. 

117.  Excitation  of  the  Field  of  Generators.  D.  C:  gen- 
erators and  motors  are  often  classified  with  regard  to  the 
manner  in  which  the  field  magnets  are  excited,  as  Separately 
Excited  or  Self  Excited.  The  Self  Excited  are  again 
divided  into  Shunt,  Series,  and  Compound  according  to  the 
manner  of  connecting  the  field  to  external  circuit.  When 
current  for  the  field  coils  of  a  generator  is  taken  from  an  out- 
side source,  the  field  is  then  said  to  be  SEPARATELY  EXCITED. 
When  the  exciting  current  is  drawn  from  the  armature 
of  the  machine  itself,  the  field  is  said  to  be  SELF  EXCITED. 

The  current  through  the  field  coils  of  a  self -excited  gen- 
erator depends  upon  the  brush  potential;  It  is  often 
desirable  to  maintain  a  field  strength  independent  of  brush 
potential,  and  separately  excited  generators  are  then  used. 
This  happens  most  frequently  in  the  testing  department 
of  a  manufacturing  plant. 


180 


ELEMENTS  OF  -ELECTRICITY 


118.  Separately  Excited  B.C.  Machines.  Self-excited 
machines  sometimes  change  their  polarity  on  starting  and 
stopping.  It  is  customary,  therefore,  to  use  a  separately 
excited  generator  whenever  it  is  desired  that  the  current 
shall  never  change  its  direction.  This  point  is  of  greatest 
importance  in  electroplating. 

Fig.     151    represents   the   connections   for   a  separately 

excited  generator  supplying 
current  to  a  set  of  electro- 
plating vats.  The  fields  are 
excited  from  a  storage  battery, 
thus  insuring  a  permanent 
polarity.  Fig.  152  shows  the 


I 


FIG.  151. — Separately  excited  generator 
supplying  current  to  electro-plating 
vats. 


FIG.  152. — Conventional  way  of  repre- 
'  excited  gt 
iting  vats. 


senting  separately  excited  generator 
feeding  electro-plat 


standard  way  of  representing  a  separately  excited   gen- 
erator, connected  to  electroplating  vats  V. 

119.  Self  Excitation.      Series- Wound.      Fig.    153  shows 
a  series-wound  generator.      Fig.  154  is  the  standard  dia- 


Fid.  153. — Series  wound  generator  feeding  series   arc  lamps. 

gram  for  the  same.     The  same  current  that  goes  through 
the  main  line  of  arc  lamps  also  goes  through  the  field  coils. 


THE  GENERATOR  181 

The  field  is  thus  in  series  with  the  external  circuit.  When 
a  series  generator  is  not  delivering  current  there  is  only 
the  residual  magnetism  in  the  fields.  Thus  the  voltage  is 
almost  zero. 

The  more  current  that  is  used  by  the  line,  the  more  highly 
magnetized  the  field  becomes  and  thus  the  higher  the  voltage 
induced  in  the  armature.    The 
external   circuit   must   there- 
fore be  arranged  to  keep  about 
the     same     current     flowing 
through  the  circuit  and  thus 
maintain    the  field  constant. 

It    WOUld     not     do     tO     have    a        FlG    154._conventional  diagram  for 

great  number  of  incandescent  ^ss  eeiierat°r  feedine  series  arc 
lamps  in  multiple  on  the  line. 

Every  time  someone  turned  off  a  set,  the  voltage  in  all  the 
rest  would  drop. 

For  this  reason  a  series  generator  is  called  a  Con- 
stant Current  Generator,  not  because  it,  of  itself, 
delivers  a  constant  current,  but  because  it  works  best 
on  a  circuit  where  a  constant  current  is  desired  and 
maintained. 

Arc  lamps  in  series  form  one  of  the  best  loads  of  this 
nature.  Several  thousand  volts  are  generated  in  order  to 
force  the  current  through  these  lamps.  When  a  lamp  is 
not  in  use,  it  is  short-circuited  so  as  not  to  break  the  main 
circuit.  When  several  lamps  are  thus  " shorted"  the  current 
(and  thus  also  the  voltage)  is  likely  to  become  excessive, 
on  account  of  the  decreased  resistance  of  the  circuit.  A  re- 
sistance (72)  may  be  introduced  to  take  the  place  of  the  idle 
lamps.  Another  device  for  controlling  the  voltage  is  to 
put  a  variable  resistance  in  multiple  with  the  field  coils, 
and  thus  make  the  amount  of  current  going  through  the 
field  coils  any  desired  fraction  of  the  current  in  the  outside 
circuit,  instead  of  all  of  it. 

Owing  to  the  high  brush  potential  of  the  commercial 


182 


ELEMENTS  OF  ELECTRICITY 


SERI 


IS  GENERATOR 


Amperes 


series  generator,  great  care  must  be  observed  in  handling 

such  a  circuit. 

If  we  plot  a  curve  between  the  amperes  delivered  by  a 

series  generator  and  the  brush  potential  while  running  at  a 

constant  speed,  we  obtain  a  curve  like  that  of  Fig.   155. 

Notice  that  the  curve  does 
not  start  from  the  origin  on 
account  of  the  fact  that  there 
must  be  always  some  mag- 
netism (residual)  in  the 
fields,  even  with  no  current 
through  the  field  coils.  If 
there  is  no  residual  magne- 
tism in  the  field,  the  ma- 
chine will  have  no  induced 
voltage  with  which  to  start  a 
small  current  flowing  through 

FIG.    155.— Relation    of  brush  potential    the  field  Coils  in  Order    to  6X- 
of  series  generator  to  current  delivered. 

cite  the  fields  Thus  the 
machine  would  fail  to  "  build  up  "  a  brush  potential. 

Note  also  that  the  voltage  increases  very  rapidly  with 
any  increase  of  the  amperes  delivered,  until  the  point  x 
is  reached.  At  that  point  the  fields  are  saturated,  so  that 
an  increase  in  current  through  the  field  coils  does  not 
thereafter  give  a  very  large  increase  in  magnetization,  and, 
therefore,  no  great  increase  in  voltage. 

The  fact  that  the  voltage  actually  falls  off  if  too  much 
current  is  taken  from  the  generator  is  due,  in  part,  to  the 
great  magnetic  disturbances  produced  in  the  armature  by 
excessive  currents  through  the  armature  windings,  and  in 
part  to  the  increased  armature  drop. 

The  field  coils  of  a  series  generator  should  be  wound  with 
a  few  turns  of  large  wire,  because  the  necessary  ampere- 
turns  must  be  obtained  with  as  little  resistance  to  the  flow 
of  the  current  as  possible.  This  is  because  the  current 
used  in  the  field  is  the  same  as  the  current  used  in  the  line, 


THE  GENERATOR 


183 


and  as  little  voltage  as  possible  should  be  used  in  forcing 
it  through  the  field  coils. 

120.  Shunt  Generator.  When  the  fields  are  excited  by 
a  current  shunted  around  the  main  circuit  as  in  Figs.  156 
and  157,  many  turns  of  fine  copper  wire  are  used.  In 
this  case  it  is  not  desirable  to  have  a  large  current  going 


Fie.  156. — Shunt  wound  generator  supplying  incandescent  lamps  in  parallel. 

through  the  coils,  because  every  bit  thus  used  is  really 
stolen  from  the  outside  circuit.  The  resistance  of  the  coils 
is  therefore  high  and  the  necessary  ampere-turns  are  ob- 
tained by  the  large  number  of  turns  which  make  up  for 
the  small  current. 

A  shunt  generator  "  builds  up  "  in  the  same  way  that 
a   series   generator   does.     The  small   amount  of  residual 
magnetism     in     the     fields 
causes  a  small  induced  volt- 
age in  the   armature.    This 
voltage  sends   a   slight  cur- 
rent through  the  field  coils, 
which    increases    the    mag- 
netization. Thus  the  induced 

-,,  -        ,  i  .      FIG.  157. — Conventional  diagram  of  shunt 

VOltage    in    the    armature    IS    generator  supplying  incandescent  lamps. 

increased.       This,    in    turn, 

increases  the  current  through  the  fields,  which  still  further 
increases  the  magnetization,  and  so  on,  until  the  saturation 
point  and  normal  voltage  of  the  machine  are  reached.  This 
"  building  up "  action  is  the  same  for  any  self-excited 


184  ELEMENTS  OF  ELECTRICITY 

generator  and  the  process  often  requires  from  20  to  30 
seconds. 

121.  Voltage  Regulation  and  Control.  A  shunt  gen- 
erator is  used  to  supply  a  circuit  where  only  fairly  constant 
voltage  is  required,  as  it  has  only  fair  VOLTAGE  REGULA- 
TION. 

By  voltage  regulation  is  meant  the  percentage  of  "  Full 
Load  "  VOLTAGE,  that  the  VOLTAGE  falls  as  the  load  rises 
from  "  no  load "  to  "  full  load  "  (the  speed  remaining 
constant) . 

no  load  voltaqe— full  load  voltage 

Voltage  Regulation  = T,  n  /    . — n -^-. 

Full  load  voltage 

By  REGULATION  is  always  meant,  some  change  which  a 
machine  makes  of  its  own  accord  when  the  load  is  changed. 
This  change  is  then  inherent  in  the  machine,  and  depends 
upon  the  construction  of  the  machine. 

By  CONTROL  is  always  meant  some  change  which  an  at- 
tendant brings  about  in  a  machine,  as  the  raising  of  the 
voltage  by  cutting  out  resistance  from  field  circuit. 

If  the  shunt  generator  is  run  at  constant  speed,  as  more 
and  more  current  is  drawn  from  the  generator,  the  voltage 
across  the  brush  falls,  though  but  slightly.  This  fall  is 
due  to  the  fact  that  it  requires  more  and  more  of  the  gen- 
erated voltage  to  force  this  increased  current  through  the 
windings  in  the  armature.  This  necessarily  leaves  a  smaller 
part  of  the  total  E.M.F.  for  brush  potential.  And  then, 
when  the  brush  potential  falls,  there  is  a  slight  decrease  in 
the  field  current,  which  depends  upon  the  brush  potential. 
This  causes  the  total  E.M.F.  to  drop  a  little,  which  still 
further  lowers  the  brush  potential. 

These  two  causes  combine  to  gradually  lower  the  brush 
potential  to  a  noticeable  degree,  especially  at  heavy  over- 
loads. The  curve  in  Fig.  158  shows  these  characteristics 
of  a  shunt  generator.  At  first,  for  small  loads,  the  curve 
runs  nearly  horizontal,  but  at  heavy  loads  it  shows  a 
decided  drop. 


THE  GENERATOR 


185 


SHUNT 


ENERAT 


The  VOLTAGE  (brush  potential)  may  be  kept  fairly  con- 
stant by  automatic  devices  for  regulating  the  speed  of  the 
''  prime  mover/'  which  may  be  a  water  wheel  or  a  steam 
engine.  By  this  means  the  speed  may  be  increasQd  whenever 
the  brush  potential  falls  a 
little,  thus  increasing  the 
total  E.M.F.  and  thereby 
raising  the  brush  potential. 

Another  method  is  to  have 
an  extra  resistance  in  the 
field  circuit,  which  may  be 
cut  out  as  the  brush  poten- 
tial falls.  This  will  allow 
more  current  to  flow  through 
the  field  coils  and  increase 
the  number  of  lines  set  up 
in  the  magnetic  circuit.  If 
the  speed  is  kept  constant, 
the  armature  conductors  cut  through  the  stronger  magnetic 
field  at  the  same  speed,  and  thus  induce  a  greater  E.M.F. 
and  restore  the  brush  potential  to  its  former  value.  This 
resistance  may  be  cut  out  either  automatically  or  by  hand. 

Example.  The  resistance  of  the  armature  of  generator  in  Fig. 
156  is  .16  ohm.  When  running  with  no  load  at  rated  speed 
the  terminal  voltage  is  130  volts,  (a)  What  will  the  terminal 
voltage  be  when  the  generator  is  delivering  50  amperes,  assuming 
field  remains  unchanged?  (6)  What  is  the  voltage  regulation 
on  this  basis? 

Armature  drop  =  50  X  .16  =  8.0  volts. 

E.M.F.  —arm.  drop  =  terminal  voltage. 

130-8  =  122  volts. 

volts  at  no  load  —volts  at  full  load 
Voltage  regulation  =  —  .  „  .  —  j— 

volts  at  full  load 

130-122 
122 


Amperes 

FIG.  158. — Relation  of  brush  potential  of 
shunt  generator  to  current  delivered. 


~122 

=  6.5%. 


186 


ELEMENTS  OF  ELECTRICITY 


Of  course  the  field  would  also  decrease,  due  to  the  current  in 
the  field  coils  becoming  less  as  the  terminal  voltage  fell.  This 
would  cause  the  terminal  voltage  to  fall  to  a  point  still  lower  than 
122  volts. 

Problem  14-7.  In  the  generator  of  Fig.  156,  the  brush  potential 
falls  from  130  volts  at  no  load  to  116  volts  at  full  load.  What  is 
the  voltage  regulation? 

Problem  15-7.  Assuming  12  volts  of  the  fall  in  voltage  in 
Problem  14-7  to  be  due  to  pressure  used  in  forcing  current  through 
armature,  what  is  the  full  load  current  of  machine?  Resistance  of 
armature  =  .16  ohm. 

Problem  16-7.  Assume  the  resistance  of  armature  in  Fig. 
150  is  .020  ohm.  Voltage  at  no  load  is  135  volts.  When  shunt 
field  alone  is  used,  the  voltage  regulation  is  7  per  cent.  What 
is  the  full-load  voltage? 

Problem  17-7.  Assuming  90  per  cent  of  the  change  in  voltage 
from  no  load  to  full  load  in  generator  of  Problem  16  is  due  to  arma- 
ture drop,  how  much  current  does  each  path  in  armature  carry 
at  full  load? 


122.  Compound  Generators.  Another  method  for  keep- 
ing the  voltage  constant  is  to  add  a  set  of  series  windings 
to  the  field,  as  in  Figs.  159,  160  and  161.  These  series  coils 


c--; 


FIG.  159. — Compound  generator  feeding  incandescent  lamps  in  parallel. 

will  tend  to  increase  the  magnetization  of  the  field  as  the 
current  increases,  just  as  the  coils  of  a  series  generator. 
This  increase  in  the  magnetization  of  the  fields  as  the 
current  is  increased  will  cause  the  voltage  to  rise  and  thus 


THE  GENERATOR  187 

compensate  for  the  fall  in  voltage  that  would  take  place  if 
the  machine  were  a  simple  shunt  generator. 

If  the  series  coils  are  adjusted  so  that  the  voltage  remains 
practically  constant  at  all  loads,  the  generator  is  said  to  be 
Flat-Compounded. 

It  is  sometimes  desirable  for  the  brush  potential  to 
actually  rise  in  order  to  make  up  for  the  increased  "  line 
drop,"  etc.,  when  the  load  increases.  In  this  case,  the 
machine  is  said  to  be  Over-Compounded.  By  this  method 
a  constant  voltage  might 
be  maintained  at  a  point 
some  distance  from  the 

R  )° 

generator     under     varying 
loads. 

There  are  two  ways  of 

Connecting  the  Shunt  field  in       FlG-  160,— Diagram  of  short  shunt  com- 
pound generator. 

compound  generators.    One 

method,  shown  in  Fig.  160,  is  to  connect  the  shunt  coils  B 

around  the  armature  A  only.     This  is  called  a  Short  Shunt. 
The  other  method,  shown  in  Fig.  161,  is  to  connect  the 

shunt  coils  B  around  both  the  armature  A  and  the  series 

coils  C.     This  is  called  a  Long  Shunt. 

If  three  generators  of  the  same  shape  and  K.  W.  output  were 

to  have  the  fields  wound  in 
the  three  different  ways, 
series,  shunt,  and  compound, 
the  number  of  ampere- 
turns  would  be  the  same 
in  each.  As  far  as  effici- 
ency goes,  one  type  has  no 
FIG.  161--Djagramnoenong  shunt  com-  advantage  over  the  others. 

The  question   of   selecting 

any  one  of  the  three  depends  upon  the  use  to  which  it  is  to 
be  put. 

When  several  shunt  or  compound  machines  are  used  in 
one  station,  as  is  very  common,  they  are  almost  always 


188  ELEMENTS  OF  ELECTRICITY 

operated  in  parallel.  Series  machines  are  operated  in 
series  whenever  used  in  combination.  This,  however,  is 
not  a  common  practice. 

Example.  Assume  that  there  are  9000  turns  in  the  shunt 
field  of  generator,  Fig.  150.  No  load  voltage  equals  120  volts. 
Current  through  shunt  field  coils  at  120  volts,  is  .8  ampere.  In 
order  that  the  full  load  voltage  equal  120  volts,  it  is  found  that 
9600  ampere-turns  are  necessary.  If  the  full  load  current  is  100 
amperes,  how  many  turns  must  be  wound  on  the  series  fields  to 
flat  compound  the  machine? 

Ampere-turns  in  shunt  field  =  9000  X. 8  =7200 
Ampere-turns  to  be  added  in  series  field  =  9600  -  7200  =  2400 

Turns  to  be  added 

10U 

=  24 

Problem  18-7.  How  many  series  turns  must  be  added  to  the 
field  of  above  example  if  the  full  load  current  is  200  amperes? 

123.  Losses  in  a  Generator.  In  Chapter  IX  it  is 
stated  that  there  are  three  sources  of  loss  in  a  generator. 

(1)  Mechanical  losses:    friction,  windage,  etc. 

(2)  Copper  loss  (PR) :  in  field  and  armature  windings. 

(3)  Iron  losses:   hysteresis  and  eddy  currents. 

The  value  of  the  mechanical  losses  may  be  obtained  by 
mechanical  means. 

The  copper  loss  we  have  seen  is  computed  by  multiply- 
ing the  square  of  the  current  in  the  armature  by  the  re- 
sistance of  the  armature  and  adding  to  that,  the  product 
of  the  square  of  the  current  in  the  field  multiplied  by  the 
resistance  of  the  field,  i.e. : 

Pa=Ia2Ra 

Pf=l?Rf 
Pa+Pf  =  Copper  loss. 


THE  GENERATOR  189 

Of  the  iron  losses  we  have  seen  that  the  part  due  to 
hysteresis  may  be  determined  from  the  equation 


107 

This  gives  the  loss  in  watts  for  each  cubic  centimeter  of 
iron  going  through  the  magnetic  cycle  once  every  second. 

It  remains  to  discuss  the  EDDY  CURRENT  Loss. 

124.  Eddy  Current  Loss.  We  have  seen  that  when  any 
conductor  cuts  lines  of  force,  there  is  always  set  up  an 
induced  electromotive  force  which  tends  to  send  an  electric 
current  along  the  conductor.  This  is  the  fundamental 
principle  underlying  the  construction  of  the  generator. 

The  copper  wires  of  the  armature  which  are  to  carry  this 
induced  current  are  wound  on  an  iron  core.  This  iron  core 
is,  itself,  an  electrical  conductor,  and  when  it  revolves  in 
a  magnetic  field,  it  cuts  lines  of  force.  This  sets  up  an 
E.M.F.  across  certain  portions  of  the  core  which  tends  to 
send  an  electric  current  through  the  core. 

Currents  thus  set  up  are  called  FOUCAULT  or  EDDY  CUR- 
RENTS, and  if  they  are  allowed  to  flow  around  in  the  iron 
they  will  soon  heat  up  the  core  and  damage  the  windings 
on  it.  Also,  it  is  always  true,  that  there  is  required  an 
expenditure  of  power  to  set  up  an  electric  current.  As  we 
cannot  use  these  eddy  currents  within  the  armature  core, 
the  power  used  in  generating  them  is  a  dead  loss.  It  is 
therefore  necessary  to  reduce  these  eddy  currents  to  as 
small  a  value  as  possible. 

This  is  done  by  laminating  the  core;  that  is,  building  it 
up  of  thin  sheets  of  iron,  insulated  from  one  another  by 
japan  or  iron  rust.  This  does  not  much  affect  the  magnetic 
circuit,  but  cuts  down  these  electric  currents,  since  the  lami- 
nations are  "  transverse  "  to  the  direction  in  which  they  tend 
to  flow.  The  thinner  the  laminations,  the  more  effective 
is  this  method  in  decreasing  the  eddy  current  loss.  Fig. 
162  is  a  photograph  of  a  pole  face  laminated  in  this  way. 


190  ELEMENTS  OF  ELECTRICITY 

There  is  no  formula  sufficiently  accurate  to  determine 
the  eddy  current  loss  in  the  core  of  a  given  generator, 
owing  largely  to  the  impossibility  of  knowing  the  insula- 
tion resistance  between  the  plates.  For  this  reason  the 
designer  of  a  generator  usually  "  estimates  "  rather  than 
"calculates  "  the  probable  eddy  current  loss  of  his  machine 
by  means  of  tables  made  up  from  data  taken  on  similar 
machines. 

It  is  well  to  remember,  however,  that  the  eddy  current 
loss  is  proportional  to  the  square  of  the  speed,  to  the  square 


FIG.  162. — Laminated  pole  piece.     Westinghouse  generator. 

of  the  flux  density,  and  to  the  square  of  the  thickness  of  the 
laminations.  When  slotted  armature  cores  are  used,  it 
is  customary  to  laminate  the  pole  faces  as  well  as  the  arma- 
ture core.  Fig.  163  explains  the  reason  for  this. 

The  teeth,  between  which  the  conductors  lie,  become  more 
densely  magnetized  than  the  slots,  and  act  like  small 
poles.  When  they  move  across  the  pole  face,  therefore, 
the  pole  face  is  really  cut  by  lines  of  force  and  small  eddy 
currents  are  set  up  as  shown  in  Fig.  163(6).  Laminating 
the  pole  face  breaks  up  the  electric  circuits  and  diminishes 
these  currents. 


THE  GENERATOR 


191 


For  a  discussion  of  the  use  of  EDDY  CURRENTS  in  damp- 
ing watt-hour   meters,  etc.,  see  Chapter  XIV.      For  their 


FIG.  163. — Action  of  slotted  armature  on  pole  face. 


effect   upon   the   efficiency   of   generators   and   motors   see 
Chapter  IX. 


192  ELEMENTS  OF  ELECTRICITY 


SUMMARY  OF  CHAPTER  VII 

ELECTROMAGNETIC  INDUCTION.  When  an  electric 
conductor  cuts  magnetic  lines  of  force,  an  E.M.F.  is  set  up 
across  the  conductor,  proportional  to  the  rate  of  cutting. 

DIRECTION  OF  INDUCED  E.M.F.  (RULE.)  If  the 
Thumb,  Forefinger,  and  Middle  Finger  of  the  Right  Hand 
are  held  at  right  angles  to  one  another,  with  the  Thumb 
extended  in  the  direction  of  the  motion  of  the  Conductor, 
the  Forefinger  in  direction  of  the  Flux,  then  the  Middle  Finger 
will  indicate  the  direction  of  the  Induced  E.M.F. 

AMOUNT  OF  E.M.F.  When  108  magnetic  lines  are  cut 
per  second,  one  volt  of  E.M.F.  is  induced. 

MODERN  GENERATORS  are,  fundamentally,  loops  of 
wire  revolved  so  as  to  cut  through  a  strong  magnetic  field.  An 
Alternating  E.M.F.  is  induced  in  these  loops,  the  curve  of 
which  approximates  a  Sine  Curve. 

If  Collecting  Rings  are  attached  to  ends  of  these  coils, 
an  alternating  current  is  delivered  by  the  Generator. 

If  a  COMMUTATOR  is  used,  a  direct  current  is  delivered. 
Since  the  armature  coils  are  each  connected  across  a  gap 
between  two  commutator  segments,  the  brushes  must  contin- 
ually short  circuit  some  two  coils. 

ANGLE  OF  LEAD.  AXIS  OF  COMMUTATION.  The 
brushes  must  be  so  placed  that  the  coils  short  circuited  by 
them  are  not  cutting  lines  of  force  at  the  instant  of  the  short 
circuit.  This  position,  owing  to  the  cross  magnetization 
due  to  the  current  in  the  armature  coils,  is  on  a  line  ahead 
of  the  neutral  axis  by  the  angle  of  lead.  This  position  is 
called  the  AXIS  OF  COMMUTATION. 

THE  NEUTRAL  AXIS  may  be  defined  as  a  line  which 
bisects  the  angle  between  the  axes  of  the  poles. 

THE   E.M.F.    INDUCED    IN    a    D.C.    ARMATURE    equals 

• — g  of  the  number  of  lines  cut  per  second  by  the  conductors 

in  series  between  any  two  adjacent  brushes;  that  is,  in  any 
single  armature  path  (there  being  as  many  paths  as  brushes.) 

THE  ARMATURE  RESISTANCE,  from  plus  terminal 
to  negative  terminal,  equals  the  resistance  of  the  wire  wound 
on  the  armature  divided  by  the  square  of  the  number  of  brushes. 

D.C.  GENERATORS  are  divided  into  two  classes:  1st, 
Separately  Excited;  2d,  Self  Excited. 


THE   GENERATOR  193 

SEPARATELY  EXCITED  Generators  have  their  fields 
excited  by  some  outside  source  of  current,  and  are  used  (ist) 
wherever  the  field  strength  must  be  independent  of  the  terminal 
voltage ;  (2d)  wherever  a  given  polarity  is  the  essential  factor. 
They  have  very  limited  use. 

SELF-EXCITED  machines  are  divided  into  three  types, 
according  to  the  method  of  connecting  the  field  coils  to  the 
receiving  circuit: 

(ist)  Series. 

(2d)  Shunt. 

(3d)  Compound. 

SERIES.  All  the  current  delivered  to  the  line  flows  through 
field  coils,  which  consist  of  a  few  turns  of  heavy  wire.  Voltage 
rises  as  load  increases.  Operates  satisfactorily  on  a  constant 
current  line. 

SHUNT.  Field  is  connected  in  parallel  with  line  and  only 
a  small  current  goes  through  coils,  which  consist  of  many  turns 
of  fine  wire.  Is  a  nearly  constant  potential  generator,  the 
voltage  falling  slightly  as  the  load  increases.  Voltage  may 
be  controlled  somewhat  by  field  rheostat. 

COMPOUND.  The  field  consists  of  two  sets  of  coils;  one 
series  and  the  other  shunt.  When  enough  series  turns  are 
wound  on,  to  offset  exactly  any  fall  in  terminal  voltage  due 
to  increased  armature  drop,  etc.,  the  machine  is  said  to  be 
Flat-Compounded.  When  terminal  voltage  rises  slightly  with 
load,  it  is  said  to  be  Over-Compounded. 

EDDY  CURRENT  LOSS.     The  losses  in  a  generator  are 
1st,    Mechanical  losses ;   (friction  and  windage.) 
2nd,  Copper  losses ;   (I2R  losses.) 
3rd,  Iron  losses ;   (hysteresis  and  eddy  currents.) 

Eddy  currents  are  local  currents  set  up  within  the  core  of 
the  armature  and  of  the  poles,  when  they  cut  magnetic  flux. 
They  cause  heating  and  consume  power;  are  greatly  dimin- 
ished by  the  use  of  laminated  parts. 

There  is  no  method  by  which  the  eddy-current  losses  in 
modern  machines  may  be  computed  to  any  degree  of  accuracy. 
The  losses  are,  accordingly,  estimated  by  comparison  with 
machines  of  similar  design  and  characteristics. 


194  ELEMENTS  OF  ELECTRICITY 


PROBLEMS     ON     CHAPTER     VII 

19-7.  An  armature  has  100  conductors  connected  in  series. 
If  these  conductors  cut  800  times  a  minute  through  a  field  of 
5,000,000  lines,  what  average  voltage  is  induced? 

20-7.  Assuming  that  the  resistance  of  the  armature  con- 
ductors in  Problem  19  is  20  ohms,  and  if  the  external  circuit  is 
zero,  what  power  is  consumed  in  the  armature  circuit? 

21-7.  Assume  the  external  resistance  to  be  100  ohms,  instead 
of  zero,  what  power  is  consumed  in  the  complete  circuit  in  Prob- 
lem 20? 

22-7.  (a)  How  much  power  is  consumed  by  the  armature 
alone  in  Problem  21-7?  (ft)  What  is  the  brush  potential? 

23-7.  Assuming  no  friction,  how  many  H.P.  would  be  required 
to  drive  armature  of  Problem  21? 


24-7.  The  drum  armature  of  a  4-pole  generator  has  a  si 
of  1200  R.P.M.  Number  of  active  conductors  =  418.  Nimiber 
of  brushes  =  4.  (/>,  per  pole,  =4,000, 000.  What  E.M.F.  does  it 
generate? 

26-7.  If  the  armature  in  Problem  24,  is  wound  with  5600  ft.  of 
No.  10.  B.  &  S.,  what  is  the  armature  resistance? 

26-7.  What  current  will  generator  in  Problem  25  deliver  when 
connected  to  an  external  circuit  of  2.4  ohms? 

27-7.  What  will  be  the  brush  potential  of  generator  in  Prob- 
lem 26? 

28-7.  How  would  you  rate  the  generator  of  Problem  27,  if  this 
is  its  full  load;  (a)  as  to  K.W.  output;  (6)  as  to  its  voltage  regula- 
tion? 

29-7.  How  much  power  is  used  in  each  armature  winding 
of  Problem  27? 

30-7.  A  ring  armature  of  a  2-pole  generator  has  480  turns,  a 
speed  of  1000  R.P.M.  <£,  per  pole,  =8,000,000  lines.  What  E.M.F. 
is  induced? 

31-7.  The  armature  of  a  bipolar  generator  has  450  conductors. 
0  =  6,000,000  lines.  Induced  E.M.F.  =  120  volts.  What  is  the 
speed  in  R.P.M.? 


THE  GENERATOR  195 

32-7.  The  armature  of  a  6-pole  generator  has  1200  active 
conductors.  The  speed  is  900  R.P.M.  Area  of  each  pole  face  is 
800  sq.cms.  Flux  density  in  air  gap  =  7500  gausses.  Number 
of  brushes  6.  What  E.M.F.  is  induced? 

33-7.  Armature  resistance  in  Problem  32  is  .12  ohm.  What 
power  is  generator  delivering  to  external  circuit  at  full  load  if  the 
brush  potential  falls  24  volts? 

34-7.  What  is  voltage  regulation  in  Problem  33? 

35-7.  If  ring  armature  of  Problem  30  were  used  in  an  8-pole, 
8-brush  generator,  what  voltage  would  be  induced,  other  data 
remaining  unchanged? 

36-7.  What  would  the  armature  resistance  be  if  armature 
of  Problem  25  were  used  in  a  6-pole  6-brush  generator? 

37-7.  What  voltage  would  be  induced  in  armature  of  Problem 
24,  if  used  on  a  6-pole,  6-brush  generator,  and  other  data  of  problem 
remained  unchanged? 

38-7.  The  resistance  of  the  field  coils  of  a  1  K.W.  shunt  generator 
is  220  ohms.  Full  load  brush  potential  is  110  volts,  (a)  What 
current  is  used  by  field  coils?  (b)  Power  consumed  in  field  coils 
equals  what? 

39-7.  (a)  How  many  turns  are  there  in  shunt  field  of  Problem  38, 
if  the  ampere  turns  equal  1000?  (6)  If  machine  were  a  series 
generator  of  same  output,  how  many  turns  would  be  required? 

40-7.  A  12-pole,  12-brush,  750-K.W.  generator,  has  a  terminal 
voltage  of  660  volts  on  full  load.  There  are  5000  ft.  of  No.  4, 
B.  &  S.  copper  wire  wound  on  armature.  What  E.M.F.  must  be 
induced  in  armature? 

41-7.  If  there  are  100  active  conductors  in  each  circuit  of  gen- 
erator in  Problem  40,  what  must  the  flux  per  pole  be?  Speed  = 
120  R.P.M. 

42-7.  How  much  would  the  flux  of  generator  in  Problem  31-7, 
have  to  be  increased  in  order  to  have  a  terminal  voltage  of  115 
volts,  when  delivering  a  current  of  20  amperes?  Armature  re- 
sistance =1.8  ohms. 

43-7.  The  total  resistance  of  a  series  generator  is  .12  ohm. 
When  delivering  10  amperes,  the  brush  potential  is  100  volts. 
Find  the  brush  potential  when  machine  runs  at  same  speed,  but 
delivers  20  amperes.  Assume  the  field  strength  increases  50  per 
cent  with  increased  current. 


196  ELEMENTS   OF   ELECTRICITY 

44-7.  In  compound  generator,  Fig.  164; 

Resistance  of  arm.  (A)  =  .04  ohm; 

"  shunt  field  (L)  =  550  ohms ; 
"series     "     (S)  =  .03ohm. 
When  delivering  60  amperes  to  line  at  550  volts,   (a)  What  is 

E.M.F.    of   generator?      (b)    How 

much    current  flows  through    ar- 
mature? 

45-7.  The   drum  armature  of  a 
bipolar   generator   is  wound   with 

600  turns  of  No.  8,  B.  &  S.  copper 

Fro.  164.— Long  shunt  compound      wire.     Each  turn  is  3.5  ft.  long. 

generator.  ^nat  is  the  armature  resistance? 

46-7.  Allowing  1000  amperes  per  sq.in.  of  cross-section  of 
conductor,  how  many  amperes  should  generator  of  Problem  45 
deliver? 

47-7.  Assuming  that  580  of  the  turns  on  the  armature  of  Problem 
46  are  active,  what  E.M.F.  must  be  generated  to  maintain  a  brush 
potential  of  110  volts  at  full  load? 

48-7.  What  must  be  the  value  of  <£  in  Problem  47?    R.P.S.=2U. 

49-7.  What  would  be  the  resistance  of  armature  and  the 
current  delivered  to  the  brushes  in  Problem  46,  if  the  armature 
were  used  in  an  8-pole,  8-brush  generator? 

60-7.  An  8-pole,  8-brush  generator  has  640  active  conductors 
on  armature  which  makes  180  R.P.M.  No  load  voltage  is  115 
volts.  Area  of  each  pole  face  is  5800  sq.cms.  What  is  the  flux 
density? 

51-7.  When  generator  in  Problem  50  is  delivering  2000  amperes 
the  brush  potential  is  110  volts.  If  the  armature  resistance  is 
.004  ohm,  what  must  the  flux  density  be  at  this  load?  Same  speed. 

52-7.  The  resistance  of  the  armature  in  a  bipolar  shunt  gen- 
erator is  .40  ohm.  The  field  resistance  is  50  ohms.  If  the  gen- 
erator delivers  40  amperes  to  the  line  at  a  brush  potential  of  100 
volts,  what  E.M.F.  must  it  generate? 

53-7.  What  current  flows  in  each  armature  circuit  of  generator 
in  Problem  52? 

54-7.  The  resistance  of  the  armature  of  a  series  generator  is 
.20  ohm;  of  the  field,  .25  ohm.  The  generator  supplies  25  arc 


THE  GENERATOR  197 

lamps,  each  of  which  takes  9.5  amperes  at  55  volts.     What  must 
be  E.M.F.  of  generator? 

55-7.  Assume  that  1  mile  of  No.  10,  B.  &.  S.,  copper  wire  is 
used  in  line  of  arc  lamps  in  Problem  54.  (a)  What  E.M.F.  must  be 
generated?  (6)  What  is  brush  potential? 

56-7.  The  resistance  of  the  armature  of  a  10-pole,  10-brush 
generator  is  .005  ohm.  <£,  per  pole,  =8,000,000  lines.  Speed  = 
180  R.P.M.  Generator  delivers  a  current  of  1200  amperes  at 
115  volts.  How  many  active  conductors  are  there  in  the  armature? 

57-7.  Each  lamp  in  Fig.  164a  takes  25   amperes  at   110  volts. 
Armature  has  .04  ohm  resistance. 
Find: 

(1)  E.M.F.  of  generator. 

(2)  Current  in  armature. 

4k 

.02  Ohms. 


=  .0».0hins. 


.02  Ohms. 


Fio.  164a. 

58-7.  Find  E.M.F.  and  armature  current  of  machine  in  Problem 
57,  when  connected  as  a  short-shunt  compound  generator  and 
supplying  same  circuit. 

59-7.  Assuming  that  the  same  power  is  lost  in  the  field  windings 
of  both  machines  in  Problem  39-7,  what  must  be  the  resistance  of 
the  series  field? 

60-7.  In  designing  a  generator  to  have  an  E.M.F.  of  230  volts, 
a  drum  armature  is  wound  with  400  conductors.  Two  brushes 
are  used  which  reduce  the  number  of  active  conductors  to  360. 
The  flux  per  pole  is  3,000,000  lines.  How  many  poles  must  gen- 
erator have  in  order  to  run  as  nearly  as  possible  at  a  speed  of  480 
R.P.M.? 

61-7.  What  would  the  flux  per  pole  in  Problem  60  have  to  be 
to  produce  exactly  the  230  volts  desired? 

62-7.  A  compound  generator,  short  shunt,  is  delivering  power 
to  the  following  pieces  in  multiple.  1000  20-o.p.,  110-volt 


198  ELEMENTS  OF   ELECTRICITY 

tungsten  lamps,  taking  1.25  watts  per  candle.  One  110-volt 
15-H.P.  motor,  with  an  efficiency  of  90  per  cent.  The  resistance 
of  shunt  field  of  generator  =  35  ohms;  of  series  field,  .007  ohm;  of 
armature,  .014  ohm;  of  line  wire,  .03  ohm. 

Find: 

(a)  Armature  current. 

(6)  Field  current. 

(c)  E.M.F. 

63-7.  What  are  the  brush  potential  and  K.W.  rating  of  gen- 
erator in  Problem  62? 

64-7.  If  a  short-shunt  generator  were  used,  what  would  be 
the  answers  to  Problems  62  and  63?  All  other  data  as  given  in 
Problem  62. 

65-7.  From  the  following  data,  find  the  number  of  ampere- 
turns  required  in  the  field  coils  of  a  115- volt  2-pole  generator. 
Armature  has  280  active  conductors.  Speed  =  1800  R.P.M. 
Magnetic  circuit  is  made  up  of  the  following  parts  in  series:  120 
cms.  of  cast  steel,  250  sq.cms.  cross-section;  20  cms.  of  wrought 
iron  240  sq.cms.  cross-section;  1.4  cms.  air  gap,  260  sq.cms.  cross- 
section.  Assume  no  leakage  of  magnetic  lines. 


CHAPTER   VIII 
MOTORS 

Force  on  Wire  in  Magnetic  Field — Torque  of  Motor — Power  Necessary 
to  Drive  Generators — Damping  of  Electrical  Instruments — Counter 
E.M.F. — Armature  Reaction  in  Motors — Direction  of  Rotation 
of  Motor— Shunt  Motor — Starting  Box — Speed  Regulation  and 
Control  of  Shunt  Motor — No  Voltage  Release — Overload  Release — 
Starting  and  Stopping  Shunt  Motor — Series  Motor:  Starting — 
Series- Parallel  Control — Compound:  Differential  and  Cumulative 
— Motor  and  Generator  Characteristics  Compared. 

IN  order  to  understand  the  operation  of  a  motor,  it  is 
necessary  to  refer  to  Figs.  27  and  28,  Chapter  II. 

Fig.  27  shows  the'  circular  magnetic  field  about  a  wire 
carrying  a  current.  Fig.  28  shows  the  effect  produced 


FIG.  165. — Field  about  a  loop  carrying  an  electric  current  in  a  magnetic  field. 

by  placing  this  wire  in  a  magnetic  field.  The  result  is 
a  condensation  of  the  lines  above  the  wire  and  a  thinning 
out  of  the  lines  below  the  wire.  The  "  rubber-band  " 
effect  of  the  lines  then  causes  a  downward  push  on  the 
wire. 

125.  Force  on  Wire  in  Magnetic  Field.     Torque.     Fig. 
165  shows  a  cross-section    of  a  single  loop  in  a  magnetic 

199 


200  ELEMENTS   OF   ELECTRICITY 

field.  If  a  current  is  sent  in  at  A  and  out  at  B,  as  marked, 
the  flux  would  be  weakened  below  A  and  above  B.  Thus 
the  loop  would  tend  to  revolve  counter-clockwise  as  marked. 
The  force  with  which  it  tends  to  revolve  is  proportional 
to  the  length  of  the  wire,  the  field  strength,  and  the  current 
in  the  wire.  This  force,  in  dynes,  may  be  computed  from 
the  following  equation: 

Im 


where  F  =  force  on  wire  in  dynes; 
1  =  current  in  amperes; 
H=  field  intensity  in  gausses; 
1=  length  of  wire  in  centimeters. 

Example.     A  wire  15  cms.  long  carrying  40  amperes,  is  in  a 
field  of  20,000  gausses.     What  is  the  force  acting  on  the  wire? 


=  —  X20,OOOX15 

=  1,200,000  dynes 
=  1225  grams 
=  2.71bs. 

Example.  Loop  in  Fig.  165  has  a  mean  diameter  of  16  cms. 
and  is  carrying  12  amperes.  Each  side  of  loop  is  20  cms.  long. 
Field  intensity  between  the  poles  is  15,000  gausses.  What  torque 
is  developed  with  the  loop  in  this  position? 


=  —  X15,OOOX20 

=  360,000  dynes,  on  each  wire 
=  368  grams 


MOTORS  201 


=  .525  ft.; 
Torque  =  force  X  arm 

—  one  of  couple  X  distance  between  them 

=  .8  IX.  525 

=  .426  Ibs.  ft.  torque. 

It  is  a  torque  built  up  in  this  way  that  causes  the  armature  of 
a  motor  to  rotate.  There  are  many  loops,  each  adding  its  share 
to  make  up  the  total  torque.  When  there  is  no  current  flowing 
in  the  loop  there  is  no  force  acting  on  the  sides  of  the  loop. 

Problem  1-8.  A  wire  20  cms.  long  lies  at  right  angles  to  the 
lines  in  a  magnetic  field.  The  force  on  the  wire  is  1.4  Ibs.  when 
25  amperes  flow  through  it.  How  strong  is  the  magnetic  field? 

Problem  2-8.  A  28-cm.  wire  is  urged  perpendicular  to  lines 
in  a  magnetic  field  of  20,000  gausses  by  a  force  of  2  Ibs.  What 
current  is  flowing  in.  wire? 

Problem  3-8.  (a)  Indicate  resultant 
field  if  poles  and  current  are  as  marked  in 
Fig.  166. 

(?>)  Indicate  direction  in  which  loop  will 
tend  to  turn. 

(c)  If  flux  is  8000  gausses,  length  of 
sides  of  loop  25  cms.  each  ;  current  through 
loop,  80  amperes.;  what  force  in  Ibs.  is  there  on  each  wire? 

Problem  4-8.  Loop  in  Problem  3  has  a  diameter  of  15  inches, 
what  torque  does  loop  possess? 

Problem  5-8.  If  the  field  strength  of  Problem  4  were  doubled, 
what  would  be  the  torque? 

126.  Power  Necessary  to  Drive  Generator.  The  fact 
that  there  is  a  force  on  a  wire  in  a  magnetic  field  when 
it  carries  a  current  explains  why  it  takes  power  to  drive 
a  generator  when  it  is  delivering  current  and  why  it  does 
not  take  power  when  the  generator  is  not  delivering  current, 
even  though  a  high  voltage  may  be  developed. 


202  ELEMENTS  OF   ELECTRICITY 

When  the  generator  is  not  delivering  current,  practically 
no  current  is  flowing  in  the  armature  windings,  thus  there 
is  no  force  to  be  overcome.  But  as  soon  as  a  current  flows 
through  the  armature,  this  force  is  present,  as  explained 
in  Fig.  165,  and  must  be  overcome.  In  other  words,  the 
torque  that  we  get  in  a  motor  when  we  send  a  current  through 
the  armature,  is  exactly  the  torque  we  have  to  overcome 
when  we  desire  to  run  the  machine  as  a  generator  and  draw 
this  amount  of  current  from  it.  This  fact  may  be  illus- 
trated as  follows: 

Suppose  we  wish  to  use  the  loop  in  Fig.  165  as  a  generator 
to  set  up  a  current  in  the  direction  marked.  By  applying 
the  right-hand  rule,  it  is  seen  that  the  direction  of  rotation 
must  be  the  reverse  of  the  direction  it  has  as  a  motor.  This 
shows  very  clearly  that  the  torque,  which  it  has  as  a  motor, 
must  be  overcome,  when  the  machine  is  run  as  a  generator 
and  supplies  the  current  to  the  line. 

Problem  6-8.  A  single  coil  armature  as  in  Fig.  166  is  delivering 
10  amperes.  Wires  are  15  cms.  long  and  the  field  strength  is  1 2,000 
gausses;  diameter  of  coil  is  9  inches.  What  torque  is  needed 
to  turn  armature  when  in  this  position  if  there  is  no  friction? 

Problem  7-8.  In  what  direction  would  coil  in  Problem  6  have 
to  turn  in  order  to  deliver  the  current  as  marked  in  Fig.  166? 

127.  Power  Lost  in  Setting  up  Eddy  Currents.     The  above 
also  explains  how  laminating  the  armature  cores  decreases 
the  eddy  current  loss,  by  decreasing  the  amount  of  current 
in  the  "  eddies."     Thus  little  torque  is  wasted  in  overcoming 
any  opposing  torque,  set  up  by  these  currents. 

128.  "  Damping  "  of  Electrical  Instruments.     Again  it 
explains   the   "  damping "   of   electrical   measuring   instru- 
ments.    In  the   Weston   D.C.   Ammeters   and   Voltmeters, 
for  instance,  the  moving  coil  is  wound  on  an  aluminum 
bobbin.     When  this  bobbin  is  moved  in  the  magnetic  field, 
currents  are  generated  in  it  by  the  cutting  of  lines  of  force. 
These  currents  then  produce  a  torque  on  the  bobbin  which 


MOTORS  203 

opposes  the  motion,  and  thus  the  tendency  to  keep  oscil- 
lating is  constantly  opposed  by  the  torque  of  the  current 
induced  in  the  bobbin. 

129.  Counter  Electro-Motive  Force.  We  have  just 
been  considering  the  "  motor-effect  "  that  is  present  in  a 
generator  when  it  is  delivering  a  current.  We  have  seen 
how  this  "  motor-effect  "  is  produced  by  the  current  flow- 
ing in  the  armature  coils  and  that  it  is  opposite  to  the 
motion  of  the  armature  of  a  generator. 

Similarly,  there  is  a  "  generator  effect  "  in  a  motor  which 
is  called  the  COUNTER  or  BACK  ELECTRO-MOTIVE  FORCE. 

This  will  be  clear  when  we  consider  that  a  motor  con- 
sists of  a  number  of  conductors  turning  in  a  magnetic 
field.  Now  if  conductors  move  in  a  magnetic  field  so  as 
to  cut  lines  of  force,  an  E.M.F.  will  be  set  up,  whether  we 
call  the  machine  a  generator,  a  motor,  or  anything  else. 

If  we  apply  the  RIGHT-HAND  RULE  to  the  motor  coil 
turning  as  in  Fig.  165,  we  find  that  there  must  be  a  voltage 
induced  in  A,  which  would  tend  to  send  a  current  OUT. 
The  current  is  actually  sent  IN  at  A  to  cause  the  coil  to  turn. 
By  applying  the  rule  to  B,  we  find  that  here,  too,  there 
is  a  voltage  induced  which  would  tend  to  'send  a  current 
IN,  while  the  current  is  really  flowing  OUT  of  B. 

This  pressure,  which  we  find  in  each  case  opposes  the 
flow  of  the  current  through  the  armature,  is  the  COUNTER 
E.M.F. 

It  is  called  the  Counter  E.M.F.  because  it  is  in  the  opposite 
direction  to  the  E.M.F.  impressed  across  the  armature  to 
cause  it  to  turn  as  a  motor. 

There  are  then  two  sources  of  pressure  in  the  windings 
of  a  motor  that  is  running:  one,  the  voltage  which  is 
impressed  on  the  terminals  from  an  outside  source;  the 
other,  the  voltage  which  is  set  up  by  the  windings  in  cutting 
lines  of  force.  These  two  pressures  are  in  opposite  direc- 
tions, and  thus  the  current  that  flows  must  be  propor- 
tional to  their  difference. 


204 


ELEMENTS  OF  ELECTRICITY 


We  might  represent  this  graphically  as  in  Fig.  167.  The 
arrow  Ex  represents  the  impressed  voltage  across  the  motor. 
The  counter  E.M.F.,  generated  when  the  motor  is  running 
would  be  represented  by  a  slightly  shorter  arrow,  Ec, 
pointing  in  the  opposite  direction.  The  voltage  Ex  tends 
to  send  a  current  down  through  the  armature,  while  the 
counter  E.M.F.  tends  to  send  a  current  up.  The  current 
will  flow  in  the  direction  of  the  greater  pressure  Ex.  The 
pressure  under  which  it  flows,  however,  will  be  the  difference 
between  the  two  forces  (Ex  —  Ec) .  This  is  analogous  to 
the  way  any  mechanical  body  behaves  when  acted  upon 


Fio.  167. — Ex  represents  impressed  E.M.F.     Ec  represents  counter  E.M.F. 

by  two  opposing  forces.  It  moves  in  the  direction  of  the 
greater  force,  and  accelerates  at  a  rate  proportional  to 
the  difference  between  the  opposing  forces. 

Thus  if  the  impressed  voltage  EX=IW  and  the  counter 
E.M.F.=  106  volts,  the  resulting  voltage,  forcing  the  current 
through  the  armature  of  the  motor,  equals  110  —  106,  or 
4  volts.  And  the  result,  as  to  current  flow,  is  just  as  though 
there  were  but  4  volts  drop  across  the  armature.  That  is, 

if  the  armature  resistance  of  this  motor  were  .5  ohm,  the 

4 
flow  of  current  would  be  —  =  8  amperes. 

.5 


Example.  The  armature  resistance  of  a  motor  is  .24  ohm. 
If  it  has  a  counter  E.M.F.  of  108.8  volts  when  running  on  a  110- 
volt  circuit,  what  current  does  it  take? 


MOTORS  205 


Effective  volts  =  110  -108.8  =  1.2; 

'-1= 

I      1'2       * 

1  =  ~7  =  0  amperes. 


Problem  8-8.  The  impressed  voltage  across  a  motor  armature 
is  115  volts;  the  counter  E.M.F.  =  112  volts;  current  =  6  amperes. 
What  is  resistance  of  armature? 

Problem  9-8.  It  is  desired  to  find  the  counter  E.M.F.  in  a 
motor  armature  at  a  certain  speed.  Armature  resistance  =  .2 
ohm.  Impressed  voltage  =  11  2  volts.  Current  =  16  amperes. 

Problem  10-8.  If  speed  in  Problem  9  were  reduced  one-half 
and  field  remained  the  same,  what  current  would  flow  in  the 
armature? 

Problem  11-8.  Suppose  armature  in  Problem  9  were  stopped 
altogether,  what  current  would  flow?  . 

130.  Reaction  in  Armatures  of  Motors.  Any  generator 
may  be  used  as  a  motor. 

Assume  the  armature  of  a  bipolar  generator  to  run  in 
a  certain  direction.  Instead  of  taking  a  current  from  it, 
suppose  we  send  the  current  through  it,  only  in  the  opposite 
direction  to  the  one  it  is  generating.  The  machine  will 
then  run  as  a  motor  and  in  the  same  direction  in  which  it 
was  being  run  as  a  generator.  That  is,  if  current  in  the 
machine  as  a  motor  flows  in  the  opposite  direction  to  its  cur- 
rent as  a  generator,  the  machine  turns  in  the  same  direction 
in  both.  Of  course,  the  change  in  direction  of  the  arma- 
ture current  will  magnetize  the  armature  core  in  the  opposite 
direction.  Thus  the  field  will  be  distorted  in  the  opposite 
direction,  and  the  axis  of  commutation  will  be  shifted  back 
behind  the  neutral  axis.  The  brushes  therefore  must  be 
shifted  to  about  this  line  in  order  to  maintain  sparkless 
commutation. 

A  comparison  of  Figs.  148  and  168  will  make  this,  clear. 
Fig.  148  represents  a  six  coil  armature  of  a  bipolar  D.C. 


206 


ELEMENTS  OF  ELECTRICITY 


generator.  Fig.  168  represents  a  six  coil  armature  of 
a  bipolar  D.C.  motor.  Notice  that  the  direction  of  rotation 
is  the  same.  The  direction  of  the  current  through  the 
coils  is  opposite.  The  field  distortion  is  ahead  in  Fig. 
148,  and  back  in  Fig.  168.  The  axis  of  commutation  cd 
of  the  generator  (Fig.  148)  is  ahead  of  the  neutral  axis.  The 
axis  of  commutation  cd  of  the  motor  is  behind  the  neutral 
axis  ab. 

The  brushes  of  the  generator  are  thus  said  to  have  a 
FORWARD  LEAD. 


FIG.  168. — Position  of  brushes  in  D.C.  motor. 

The  brushes  of  the  motor  (Fig.  168)  have  a  BACKWARD 
LEAD. 

The  brushes  of  the  generator  in  Fig.  148,  it  was  said,  are 
set  a  little  forward  even  of  cd.  This  is  in  order  to  bring 
the  current  in  the  shorted  coil  to  zero  as  soon  as  possible, 
by  causing  a  slight  voltage  to  be  set  up  in  it  in  the  opposite 
direction  to  which  this  current  is  flowing.  For  a  similar 
reason  the  brushes  of  the  motor  are  set  a  little  behind  cd. 

Consider  coil  No.  6  shorted,  at  a  certain  instant,  by  the 
negative  brush  Bl  of  the  motor.  As  it  reaches  the  brush 
there  must  be  a  current  flowing  in  it  in  the  direction  of 
the  current  in  coil  No.  5.  In  order  that  this  current  may 
stop  flowing  while  the  brush  shorts  the  coil,  the  brush  is 
made  to  short  it  just  before  it  moves  to  a  neutral  spot  and 


MOTORS  207 

ceases  to  cut  lines.  Of  course  the  cutting  of  the  lines  in  a 
motor  always  sets  up  a  back  E.M.F.  which  opposes  the 
current  actually  flowing.  So  the  E.M.F.  now  set  up  in 
this  shorted  coil  No.  6  opposes  the  current  that  is  flowing 
in  it  and  quickly  brings  it  to  zero.  Before  this  E.M.F.  has 
succeeded  in  actually  setting  up  a  current  in  the  opposite 
direction  the  brush  has  ceased  to  short-circuit  the  coil 
and  therefore  there  is  no  current  to  be  broken.  Thus  no 
sparking  results  as  the  segment  passes  from  under  the  brush. 

131.  Rule  for  Direction  of  Rotation  of  Motors.  When 
all  other  conditions  are  the  same,  the  current  in  a  motor 
flows  in  the  opposite  direction  to  the  current  in  a  generator. 
In  other  words,  the  back  voltage,  the  (voltage  which  the 
machine  produces  as  a  generator)  is  opposite  to  the 
impressed  voltage — the  voltage  which  causes  it  to  run  as 
c.  motor. 

The  rule  for  finding  the  direction  of  rotation  of  a  motor 
is  the  same  as  that  for  finding  the  direction  of  rotation  of 
a  generator.  We  must,  however,  deal  with  the  back 
voltage,  which  opposes  the  impressed  voltage. 

Therefore,  point  the  forefinger  of  the  right  hand  in  the 
direction  of  the  flux,  the  middle  finger  OPPOSITE  the  direction 
of  impressed  voltage,  and  the  thumb  will  point  the  direction 
of  rotation. 

NOTICE    THAT    ALL    RULES    FOR     DIRECTION    OF    MAGNETIC 

, 

FIELD,  CURRENT,  INDUCED  VOLTAGE,  MOTION,  ETC.,  APPLY 
ONLY  TO  THE  RlGHT  HAND. 

To  change  the  direction  of  rotation  of  a  motor,  it  is 
necessary  to  change  the  relation  between  the  direction  of 
the  flux  and  the  direction  of  the  armature  current.  If 
we  change  the  direction  of  both  the  flux  and  the  armature 
current  we  do  not  change  the  relation  between  them  and 
therefore  the  direction  of  rotation  remains  as  before. 

The  rule,  then,  for  reversing  the  direction  of  rotation  of 
a  motor,  is  to  reverse  either  the  field  or  armature  connections, 
not  both. 


208  ELEMENTS  OF   ELECTRICITY 

132.  The  Shunt  Motor.     There  are  three  general  types 
of  motors,  as  there  are  three  general  types  of  generators, 
SHUNT,  SERIES  and  COMPOUND.     They  are  named  accord- 
ing to  the  way  the  fields  are  connected.     The  shunt  motor 
is  by  far  the  most  common  type  of  the  three  and  will  be 
described  first. 

133.  Starting    Box.     We    have    seen    that   the    current 
which  a  motor  takes  in  the  armature  depends  upon  the 
difference  between  the  impressed  voltage  and  the  counter 
E.M.F.     This  counter  E.M.F.  keeps  the  current  from  being 
excessive   although   the   armature   resistance   is   very   low. 
We  have  seen  that  this  counter  E.M.F.  depends  upon  the 
speed,  since  it  is  caused  by  the  armature  conductors  cut- 
ting lines  of  force. 

When,  therefore,  the  armature  is  not  rotating,  there  is 
no  counter  E.M.F.  and  the  current  through  the  armature 
depends  upon  the  impressed  voltage  and  the  armature 
resistance  only.  Suppose  then,  that  full  voltage  were 
thrown  on  to  the  armature.  Since  the  resistance  is  small, 
the  current  through  the  armature  would  be  excessive  and 
might  burn  out  the  windings. 

Example.  The  armature  of  a  shunt  motor  contains  .2  ohm 
resistance.  The  motor  is  to  run  on  a  110  volt  circuit,  (a)  Sup- 
pose that  it  is  thrown  on  the  circuit  suddenly  while  the  armature 
is  standiifg  still,  what  current  will  it  take? 

/  =  — —  =  550  amperes. 

With  the  machine  running  at  normal  speed  on  110  volts  there 
is  a  Counter  E.M.F.  of  107  volts,  (ft)  What  current  does  it  take 
at  this  speed? 

_     110-107 

/  =      — - —   =  J  5  amperes. 

In  order  to  avoid  this  excessive  current  on  starting,  a 
STARTING  RESISTANCE  is  introduced  into  the  circuit  which 


MOTORS  209 

cuts  down  the  voltage  across  the  motor  armature  at  first, 
and  allows  but  a  small  current  to  flow  through  it.  By 
slowly  cutting  out  this  resistance  as  the  motor  speeds  up 
and  sets  up  its  counter  E.M.F.,  it  is  possible  to  throw  the 
full  voltage  across  the  motor.  It  is  safe  to  have  this  high 
voltage  across  the  motor  as  long  as  the  speed  is  high  enough 
to  oppose  it  by  a  high  counter  voltage,  but  at  low  speed 
(and  thus  a  small  counter  voltage)  too  much  current  is 
likely  to  be  forced  through  the  armature. 

Fig.  169  is  a  simple  diagram  of  the  starting  resistance  used 
with  a  shunt  motor. 


C 
FIG.  169. — Diagram  of  starting  resistance  for  shunt  motor. 

When  the  line  switch  is  thrown  and  C  is  swung  to  first 
point,  it  merely  puts  the  shunt  field  F  on  to  the  circuit, 
thus  building  up  the  field  immediately.  Then  when  the 
arm  C  is  swung  to  the  next  contact  point,  the  starting 
resistance  SR  is  put  in  series  with  the  armature  across  the 
line.  The  resistance  SR  prevents  too  large  a  current  from 
entering  the  armature.  As  soon  as  some  speed  is  acquired 
by  the  armature  (and  therefore  a  counter  E.M.F.)  the  arm 
C  is  swung  to  the  next  contact  point,  cutting  out  some  of 
the  resistance.  As  the  motor  gets  up  its  full  speed  the 
rest  of  the  resistance  SR  is  gradually  cut  out.  Finally  the 
armature  is  put  directly  across  the  line,  by  swinging  the 
arm  C  to  the  point  P. 

134.  Speed  Regulation  of  Shunt  Motor.  A  shunt  motor 
has  the  advantage  of  having  a  definite  speed  at  "  no  load;" 
that  is,  it  does  not  tend  to  "  run  away  "  or  "  race  "  when 


210  ELEMENTS  OF  ELECTRICITY 

the  load  is  taken  off.  The  "  full  load  "  speed  is  but  slightly 
lower  than  the  "  no  load  "  speed.  Thus  the  motor  pos- 
sesses a  good  "  speed  regulation." 

The  percentage  of  the  full  load  speed,  that  the  speed 
changes  of  its  own  accord  as  the  load  changes  from  "no 
load  "  to  "  full  load  "  is  called  the  "  SPEED  REGULATION." 

No  Load  SPEED  —  Full  Load  SPEED 

SPEED  REGULATION  =  -  —  . 

lull  Load  SPEED 

Thus  if  the  speed  of  a  machine  fell  from  1280  R.P.M. 
at  no  load  to  1200  R.P.M.  at  full  load,  the  regulation  would 


By  regulation  is  meant  always  some  change  which  the 
machine  itself  makes  as  the  load  is  changed.  By  control 
is  meant  some  change  that  an  attendant  brings  about  on 
the  machine,  as  by  throwing  of  more  resistance  into  the 
field,  to  raise  the  speed.  See  Chapter  VII. 

135.  Speed  Control  of  Shunt  Motor.  The  usual  method 
of  changing  the  speed  of  a  shunt  motor  is  to  change  the 
strength  of  the  field. 

Suppose  by  means  of  a  variable  resistance  in  the  shunt 
field,  we  increase  the  resistance  of  the  field  coils.  This 
causes  a  decrease  of  current  through  the  coils,  which  in 
turn  causes  a  decrease  in  the  magnetization.  Then  fewer 
number  of  lines  are  cut  per  second  by  the  armature  coils, 
and  thus  the  counter  E.M.F.  is  lowered.  This  allows 
more  current  to  flow  into  the  armature  coils. 

This  increase  of  current  in  the  armature  coils  is  much 
greater  in  proportion  than  the  decrease  in  the  strength 
of  the  magnetic  field.  Thus  the  actual  force  or  the  torque 
of  the  motor  is  greater,  and  the  armature  speeds  up  until 
a  balance  is  again  obtained.  Of  course,  if  the  current 
increased  only  in  proportion  as  the  field  decreased,  the 
product  of  current  times  field  strength  (which  is  always 


MOTORS  211 

the  force  on  the  armature  conductors)  would  remain  con- 
stant and  there  would  be  no  gain  in  torque,  hence  no  increase 
in  speed  when  the  field  strength  was  decreased. 

Therefore,  strange  as  it  appears  at  first  sight,  the  weaker 
the-  field,  the  faster  a  shunt  motor  tends  to  run.  This 
method  of  changing  the  field  strength  by  means  of  a  vari- 
able resistance  in  series  with  it  is  very  common. 

When  a  slower  speed  is  desired,  some  resistance  is  cut 
out  and  more  current  allowed  to  flow  through  the  field 
coils.  The  armature  coils  then  have  to  cut  through  a 
denser  field,  and  a  higher  counter  E.M.F.  is  developed, 
which  cuts  down  the  current, in  the  armature  and  causes 
it  to  run  at  a  lower  speed.  The  machine  will  keep  approx- 
imately this  speed  at  any  load  within  the  range  of  its  rating. 

There  are  limits  to  the  changes  in  speed  which  can  be 
effected  by  this  means.  (1)  It  is  not  feasible  to  increase 
the  flux  density  of  the  iron  much  beyond  saturation,  and 
thus  the  machine  can  be  slowed  down  to  a  certain  point 
only.  (2)  On  the  other  hand,  when  the  flux  density  gets 
low  in  the  field,  the  magnetic  disturbances  caused  by  the 
armature  are  so  great  that  serious  sparking  at  the  brushes 
results. 

Another  scheme  which  takes  advantage  of  this  change 
in  field  strength  to  bring  about  a  change  in  speed,  is  that 
used  by  the  Stow  Mfg.  Co.,  illustrated  by  Fig.  170. 

The  cores  of  the  fields  are  made  of  plungers  which  can  be 
drawn  away  from  the  armature,  thus  increasing  the  air 
gap.  This  of  course,  increases  the  reluctance  of  the  mag- 
netic circuit  and  thus  the  flux  density  decreases,  and  an 
increase  of  armature  speed  results. 

By  this  method  most  of  the  sparking  at  high  speeds 
is  avoided.  The  apparatus,  however,  for  changing  the 
position  of  the  plungers  in  the  field  coils  is  complicated 
and  expensive.  A  motor  equipped  with  either  device  is 
called  a  "  VARIABLE  SPEED  MOTOR." 

A  somewhat  simpler  scheme  which  allows  a  wide  range 


I 


212 


ELEMENTS   OF   ELECTRICITY 


of  speeds,  is  the  COMMUTATING  POLE,  sometimes  called 
the  INTER-POLE.  The  arrangement  of  the  field  of  this 
motor  is  shown  in  Fig.  179.  On  the  small  commutating 
poles,  placed  between  the  regular  poles,  the  coils  are  in 
series  with  the  armature.  The  strength  of  these  extra 
poles  then  depends  upon  the  current  which  the  motor 
is  taking.  Their  action  is  to  neutralize  the  excessive 


FIG.  170. — Stow  variable  speed  motor,  two  pole. 

magnetic  disturbances  which  take  place  when  the  field 
is  considerably  weakened,  in  order  to  obtain  high  speed. 
This  method  is  so  successful  in  maintaining  the  proper 
shape  of  field  that  sparkless  commutation  is  possible  at 
speeds  varying  600  per  cent  from  slowest  to  fastest. 

136.  "  No  Field "  Release.  What  would  happen  if 
the  current  in  the  field  coils  were  reduced  to  zero,  and  only 
the  residual  magnetism  were  bft  in  the  fields? 


MOTORS 


213 


We  have  seen  that  the  armature  increases  in  speed  as 
the  field  decreases  in  strength.  We  would  thus  expect  the 
armature  speed  to  become  excessive  if  the  field  became  very 
weak.  And  this  is  just  what  happens  to  a  shunt  motor 
when  the  field  is  broken,  and  the  load  is  light.  It  imme- 
diately races  and  destroys  the  armature  windings;  the 
centrifugal  force  pulling  them  away  from  the  core.  Of 
course,  if  the  machine  is  heavily  loaded,  it  stops  and  the 
greatly  increased  current  in  the  armature  coils  causes  them 
to  burn  up.  This  event  of  "  no  field  "  must  therefore  be 
guarded  against  by  some  device,  which  will  break  the  arma- 
ture circuit  automatically  as  soon  as  the  field  circuit 
is  broken,  and  thus  avoid  the  racing  or  burning  of  the  arma- 
ture coils. 

Fig.  171  shows  a  device  of  this  kind.  The  field  is  led 
through  a  small  electro-magnet  M  on  the  starting  box 


FIG.  171.— "No-field"  release. 

The  swinging  arm  Chas  a  soft  iron  keeper  K  attached  to  it. 
When  the  arm  has  come  into  the  running  position  and  all 
the  starting  resistance  SR  is  cut  out,  the  keeper  K  comes 
in  contact  with  the  electro-magnet  M  which  holds  the  arm 
in  this  position,  acting  against  the  tension  in  the  spring  S. 
If  anything  happens  to  break  the  current  in  the  field 
coils  F,  the  current  in  the  electro-magnet  is  also  broken, 
and  the  swinging  arm  is  released  and  pulled  away  by  the 
spring  S.  This  action  breaks  the  armature  circuit  and 
thus  stops  the  motor. 


214 


ELEMENTS   OF   ELECTRICITY 


137.  "  No  Voltage  "  Release.  There  is  also  sometimes 
danger  that  the  voltage  will  go  off  the  line,  and,  a  few  min- 
utes later,  will  be  thrown  on  again.  In  the  meantime,  the 
motor  will  have  slowed  down  and  possibly  stopped.  If 


FIG.  172. — "No-voltage"  release. 


the  voltage  is  now  thrown  on  with  the  arm  in  the  running 
position,  with  all  the  starting  resistance  cut  out,  the  arma- 
ture might  be  burned  out.  The  above  "  no  field  "  release 
is  designed  to  work 'also  as  a  "no  voltage"  release;  that 


FIG.  172a. — Cutler  Hammer  starting  box.      "  No-voltage  "  release.     ' 

is,  to  release  the  arm  and  throw  the  motor  off  the  line  if 
the  voltage  drops. 

It  would  appear  that  if  the  line  voltage  were  shut  off  there 
would  be  no  current  through  the  electro-magnet  M  and  the 
arm  C,  Fig.  171,  would  be  released.  The  motion  of  the 
armature,  however,  is  always  setting  up  a  counter  E.M.F. 


MOTORS 


215 


which  will  send  a  current  through  the  fields  and  electro- 
magnet M  as  long  as  it  is  turning  fast  enough.  Thus 
this  device  doesn't  act  immediately  as  a  "  no-voltage  " 
release,  though  it  will  do  so,  if  the  line  remains  "  dead  " 
long  enough  for  the  motor  to  slow  down  to  a  point  where 
the  counter  E.M.F.  will  not  excite  the  electro-magnet  M. 

No-voltage  releases  are  often  arranged  as  in  Fig.  172. 
Fig.  172a  shows  the  construction  and  appearance  of  such 
a  box.  For  diagram  of  connections  see  Fig.  229. 

138.  "  Overload "  Release.  There  must  also  be  some 
arrangement  to  prevent  putting  too  much  load  on  a  motor 


FIG.  173.— "Overload  release. 


FIG.  173a. — Ward  Leonard  starting  box  with 
"Overload"  and  "no-voltage"  releases. 


and  thus  causing  the  speed  to  become  so  slow  that  the 
armature  current  is  excessive.  An  OVERLOAD  RELEASE 
as  shown  in  Fig.  173  takes  care  of  this  emergency.  A  coil 
L  of  low  resistance  is  placed  in  the  motor  line  so  that  all 
the  current  taken  by  the  motor  must  pass  through  it. 
When  the  current  in  the  motor  becomes  excessive,  the 
coil  becomes  so  strongly  magnetized  that  it  sucks  up  the 
plunger  S  and  short  circuits  the  magnet  coil  M.  This 
destroys  the  magnetizing  force  of  M.  The  arm  C  is  thus 
released  and  the  current  shut  off  from  motor  armature  and 
field.  Fig.  173a  shows  the  appearance  of  this  box.  For 
diagram  of  connections  see  Fig.  230. 


216  ELEMENTS  OF  ELECTRICITY 

Problem  12-8.  The  line  voltage  is  115  volts.  The  resistance 
of  the  shunt  field,  F,  Fig.  171,  is  200  ohms;  of  magnet  coil  3/, 
20  ohms.  In  starting  resistance  SR  the  resistance  from  1 
to  2  is  5  ohms;  from  2  to  3,  4  ohms;  from  3  to  4,  3  ohms; 
from  4  to  5,  1  ohm;  armature  resistance  is  2  ohms.  The 
switch  to  power  is  thrown  and  contact  c  is  swung  to  point  No. 
1.  (a)  How  many  amperes  flow  through  the  armature?  (6)  How 
many  amperes  flow  through  the  field? 

Problem  13-8.  When  the  armature  of  Problem  12  has  attained 
enough  speed  to  set  up  a  counter  E.M.F.  of  25  volts,  the  contact 
c  is  swung  to  point  2.  (a)  How  many  amperes  flow  through 
the  armature?  (6)  How  many  amperes  flow  through  the  field? 

Problem  14-8.  When  the  counter  E.M.F-  of  armature  in  Problem 
13  is  50  volts,  contact  c  is  swung  to  point  3.  (a)  How  many 
amperes  flow  through  the  armature?  (6)  How  many  amperes 
flow  through  the  field? 

Problem  15-8.  The  armature  of  Problem  14  attains  a  counter 
E.M.F.  of  75  volts  and  contact  c  is  swung  to  point  4.  Answer 
(a)  and  (6)  of  Problem  14. 

Problem  16-8.  Armature  of  Problem  15  attains  a  counter 
E.M.F.  of  107  volts  and  contact  c  is  swung  to  5.  Answer  (a)  and 
(6)  of  Problem  14. 

139.  Directions    for    Starting    and    Stopping    a    Shunt 
Motor.     Start  slowly,  moving  the  swinging  arm  slowly  until 
the  motor  comes  up  to  speed.     Do  not  leave  it  on  any 
intermediate  point  for  any  length  of  time,  or  the  starting 
resistance  may  be  burned  out. 

Always  stop  the  motor  by  pulling  the  line  switch  Fig. 
171,  and  allowing  the  arm  C  to  snap  back  when  the  electro- 
magnet releases  it.  Never  pull  the  arm  C  back,  as  this 
causes  bad  arcing  across  the  points  1  and  0  which  roughens 
the  copper  contacts. 

140.  Series    Motors.       No-load    Speed.     Series    motors, 
unlike  shunt  motors,  have  not  a  definite  "  no-load  "  speed. . 
When  unloaded,  a  series  motor  will  continue  to  increase 
in  speed,  until  it  destroys  the  armature  by  the  great  cen- 
trifugal force  set  up. 


MOTORS  217 

This  may  be  seen  by  considering  the  behavior  of  the 
field  in  an  unloaded  series  motor.  The  motor  starts  up 
as  the  current  is  sent  into  the  armature  and  field  coils. 
There  being  no  opposing  torque,  since  the  motor  is 
unloaded,  the  speed  tends  to  increase  until  the  counter 
E.M.F.  equals  the  impressed  voltage.  But  the  ever  increas- 
ing counter  E.M.F.  decreases  the  current  in  the  field  and 
armature.  As  the  field  weakens,  it  requires  an  increased 
speed  to  set  up  the  desired  counter  E.M.F.  The  field  con- 
tinues weakening  and  the  speed  continues  increasing  in 
an  effort  to  build  up  a  counter  E.M.F.  equal  to  the  impressed 
-voltage,  until  the  centrifugal  force  wrecks  the  machine. 
Thus  a  series  motor  is  used  only  where  the  load  is  always 
attached  to  the  armature,  as  in  a  fan,  a  railway  car,  an 
electric  crane,  etc. 

We  have  seen  that  an  unloaded  shunt  motor  tends  also 
to  speed  up  till  its  counter  E.M.F.  becomes  equal  to  the 
impressed  voltage.  But  since  the  field  remains  constant, 
this  equality  is  approximately  reached  and  maintained  at 
a  certain  definite  speed. 

141.  Comparison  of  Shunt  and  Series  Motors.  Starting 
Torque.  We  have  seen  that  the  force  on  the  armature 
conductors,  and  therefore  the  torque,  of  any  motor  is  pro- 
portional to  the  current  /  in  the  armature  times  the  field 
strength  H.  Other  things  remaining  constant,  we  may 
write  the  proportion: 

Torque  oc  ///. 

But  in  a  series  motor,  increasing  the  current  I  in  the 
armature  increases  the  current  in  the  field  exactly  as  much, 
and  thus  increases  the  field  strength  H  nearly  as  much. 

The  magnetization  curve  approximates  a  straight  line  until 
the  saturation  point  is  reached. 

Thus  we  may  write  the  proportion  for  the  series  motor 
Torque  oc  P. 


218  ELEMENTS   OF    ELECTRICITY 

So,  if  at  any  time  the  current  in  the  armature  of  a  scries 
motor  be  doubled,  the  field  strength  is  also  doubled,  pro- 
viding saturation  is  not  exceeded,  and  the  torque  is 
increased  four  times. 

In  the  case  of  a  shunt  motor,  the  field  current  remains 
practically  constant,  so. that  doubling  the  current  means 
merely  doubling  the  armature  current.  The  torque  is  then 
only  doubled. 

In  other  words,  the  proportion  for  a  shunt  motor  is: 

Torque  oc  /. 

Accordingly,  when,  as  in  starting  a  motor,  we  allow  a 
heavy  current  to  flow  momentarily  through  the  machine, 
if  it  is  a  series  motor,  the  torque  is  proportional  to  the 
square  of  the  current.  In  the  case,  however,  of  a  shunt 
motor,  the  torque  is  proportional  to  the  current  only. 

Thus,  of  two  motors  of  same  size,  the  scries  has  a  much 
greater  starting  torque.  In  order  that  the  shunt  motor 
have  the  same  starting  torque  as  the  series  motor,  it  would 
be  necessary  to  force  so  much  current  through  the  armature 
that  it  would  be  burned  up.  In  other  words,  if  the  two 
motors  are  to  have  the  same  starting  torque,  the  shunt 
motor  must  be  built  much  heavier,  to  stand  larger  currents. 

142.  Torque-Speed  Characteristics.  The  torque  of  a 
series  motor  varies  greatly  with  the  speed,  being  greatest 
when  the  speed  is  slowest.  As  explained  above,  this  is 
because  the  slow  speed  sets  up  a  small  counter  E.M.F. 
and  the  impressed  voltage  can  force  a  large  current  through 
both  armature  and  field.  Similarly,  a  high  speed  reduces  the 
current  in  both  armature  and  field,  and  cuts  down  the  torque. 
So  when  a  heavy  load  is  thrown  on  a  series  motor,  the  speed 
decreases,  and  the  armature  current  and  field  strength  both 
increase,  with  the  result  of  a 'great  gain  in  torque,  to  take 
care  of  the  increased  load. 

On  the  other  hand,  we  have  seen  that  the  field  being 
constant,  shunt  motors  have  but  little  change  of  speed 


MOTORS  219 

with  a  variation  of  load  from  no  load  to  full  load.  So,  when 
a  heavy  load  is  thrown  on  a  shunt  motor,  it  has  to  gain 
its  increased  torque  entirely  by  the  increased  current  in 
the  armature.  In  order  to  take  care  of  the  same  load, 
therefore,  the  shunt  motor  would  have  to  be  much  larger 
than  the  series  motor. 

The  shunt  motor  cannot  slow  down  and  take  the  heavy 
load  at  a  slower  rate,  but  must  rush  through  the  hard 
job  at  about  the  same  speed  it  maintains  on  a  light 
load. 

Wherever  a  fairly  constant  speed  is  desired,  a  shunt 
motor  can  be  used,  care  being  taken  to  get  one  large  enough 
to  take  care  of  the  heaviest  load. 

When  the  speed  is  not  an  important  factor,  but  high 
torque  is  required,  a  series  motor  is  preferable.  Take 
for  instance  the  most  common  use  of  a  series  motor,  electric 
railway  work.  If  a  shunt  motor  were  used,  the  car  would 
have  to  climb  the  hills  at  about  the  s#me  rate  as  it  goes 
along  a  level,  because  the  shunt  motor  is  very  nearly  a 
constant  speed  machine.  This  would  require  a  heavy  cur- 
rent to  do  work  at  this  rate  and  would  burn  out  the  motor 
on  a  long  hill  unless  the  motor  were  several  times  the  size  of 
the  series  motor  used.  The  series  motor,  however,  decreases 
its  speed  and  thus  works  at  a  slower  rate  and  still  gains 
torque  to  be  used  in  overcoming  the  large  opposing  torque 
offered  by  the  grade. 

On  the  other  hand,  a  series  motor  would  not  do  to  rim 
a  lathe  in  a  machine  shop.  Every  time  the  load  changed, 
the  speed  would  change,  and  the  most  efficient  cutting 
speed  could  not  be  maintained.  Moreover,  as  soon  as  the 
load  was  taken  off,  the  speed  would  become  so  great  as  to 
wreck  the  lathe. 

143.  Starting  a  Series  Motor.  Series  motors  are 
started  by  placing  a  resistance  in  series  with  them  and 
gradually  cutting  it  out,  as  motor  gets  up  speed.  This 
starting  resistance  serves  the  same  purpose  as  the  starting 


220 


ELEMENTS  OF   ELECTRICITY 


resistance  which  is  placed  in  series  with  the  armature  of 
a  shunt  motor. 

144.  Series-Parallel    Control    for    Electric    Cars.      Most 
electric  cars  have  at  least  two  motors.     The  controller  shown 

in  Fig.  174  is  in  the  front  of 
the  car  and  is  operated  by 
the  motorman  as  follows: 

When  the  controller  handle 
is  advanced  to  the  first  notch, 
it  places  the  two  motors  A 
and  B  in  series  with  each 
other  and  with  the  starting 
resistance  SR  as  in  Fig.  175. 
As  the  handle  is  advanced,  it 
gradually  cuts  out  the  start- 
ing resistance  until  the  car 
gets  up  a  speed  of  about  10 
miles  an  hour  on  the  level. 
Then  the  next  notch  puts  the 
two  motors  in  parallel  with 
each  other  and  again  in 
series  with  the  resistance  SR 
as  in  Fig.  176.  If  greater 

speed  is  desired,  the  resistance  is  again  cut  out  by  advancing 
the  handle. 

The  scheme  of  putting  the  two  motors  in  series  at  the 
start,  allows  the  car  to  be  started  on  half  the  current  it 
would  take  to  start  with  them  in  parallel  and  thus  pre- 
serves a  more  even  distribution  of  current  in  the  trolley 
system,  and  wastes  much  less  power.  It  really  makes  one 
motor  act  as  starting  resistance  for  the  other,  at  the  same 
time  helping  it  to  supply  tractive  effort. 

145.  Caution  in  use  of  Series  and  Shunt  Motors.  A 
shunt  motor  races  when  the  field  is  broken,  if  the  armature 
circuit  is  not  also  broken.  Therefore: 

NEVER  PULL  THE  FIELD  OF  A  SHUNT  MOTOR. 


FIG.  IV-i. — \Vestinghouse  controller. 


MOTORS 


221 


A  series  motor    races  when  there  is  no   load  connected 
to  it.     Therefore: 

NEVER  START  AN  UNLOADED  SERIES  MOTOR  AND  NEVER 
REMOVE  ALL  THE  LOAD  FROM  A  SERIES  MOTOR  WHILE 
IT  is  RUNNING. 

146.  Compound   Motors.     Differential   and   Cumulative. 

When    absolutely    constant  speed  is  desired    under    wide 


Trolley 


Trolley 


S.R, 


S.R, 


FIG.  175. — Motors  in  series  at 
start. 


FIG.  176. — Motors  in  parallel  when 
running. 


changes  in  load,  as  in  a  machine  shop;  a  DIFFERENTIAL 
COMPOUND  MOTOR  is  sometimes  used.  This  has  a  set  of 
series  field  coils  "  bucking  "  the  shunt  coils.  The  ampere- 
turns  of  the  series  coils  D,  Fig.  177,  oppose  the  ampere- 
turns  on  the  shunt  coils  S.  The  field  strength  due  to  these 
series  coils  increases  as  the  load  increases  and  acting  against 
the  field  due  to  the  shunt  coils,  weakens  the  total  field 
enough  to  allow  an  extra  current  to  flow  through  the  arma- 
ture. This  current  will  create  enough  torque  to  overcome 
the  extra  drag  on  the  pulley  and  keep  the  speed  constant. 


222  ELEMENTS  OF   ELECTRICITY 

Of  course  the  shunt  field 'S  is  much  stronger  than  the  series 
field  D.  It,  however,  is  possible,  by  making  the  scries 
field  large  enough,  to  cause  the  speed  to  increase  with  the 
load.  In  the  main,  shunt  motors  furnish  for  most  purposes 
a  speed  which  is  constant  enough  at  all  loads. 

A  DIFFERENTIAL  COMPOUND  motor  then  has  exaggerated 
shunt  characteristics;  the  starting  torque  is  comparatively 
low  (lower  even  than  the  starting  torque  of  a  shunt  motor) ; 
the  speed  regulation  is  exceedingly  .good  (being  better 
than  the  speed  regulation  of  a  shunt  motor). 

As  the  starting  torque  of  a  series  motor  is  greater  than 
that  of  a  shunt  motor,  series  coils  are  sometimes  added 
which  "  aid  "  the  shunt  coils  instead  of  "  buck  "  them. 


FIG.  177. — Differential  compound  FIG.    178. — Cumulative   compound 

motor.  motor 


The  motor  is  then  called  a  CUMULATIVE  COMPOUND  MOTOR. 
The  ampere-turns  of  the  series  coils  C,  Fig.  178,  aid  the 
ampere-turns  on  the  shunt  coils  S.  Such  an  arrangement 
gives  the  motor  the  advantage  of  the  large  starting  torque 
of  a  series  motor,  without  the  disadvantage  of  the  machine 
•"  racing  "  on  no-load,  since  the  shunt  coils  maintain  the 
field.  The  characteristics  are  those  of  the  series  motor 
exaggerated;  the  speed  slowing  down  as  the  load  increases, 
more  than  does  the  speed  of  a  straight  series  motor.  To 
avoid  this,  the  series  field  is  often  short  circuited,  after 
the  motor  has  been  started.  The  motor  then  starts  as 
a  series  motor  but  runs  as  a  shunt  motor. 


MOTORS 


223 


147.  Comparison   between    Generators   and   Motors.     It 

is  to  be  noted  that  generators  and  motors  of  the  same  type 
have  very  similar  characteristics.     We  have  merely  to  read 


FIG.  179. — Westinghouse  commutating  pole  motor. 

speed  in  motors  for  voltage  in  generators.     The  following 
table  is  intended  to  bring  out  this  relation: 

Shunt 

GENERATORS.  Constant  voltage;  good  voltage  regula- 
tion. Voltage  controlled  by  field  variation. 

MOTORS.  Constant  speed;  good  speed  regulation.  Speed 
controlled  by  field  variation. 


224  ELEMENTS  OF  ELECTRICITY 

Series 

GENERATORS.     Voltage  depends  entirely  upon  load. 
MOTORS.     Speed  depends  entirely  upon  load. 

Compound 

GENERATORS.  Voltage  may  be  made  absolutely  uniform 
throughout  wide  range  of  loads. 

MOTORS.  (Differential).  Speed  may  be  made  absolutely 
uniform  throughout  wide  range  of  loads. 


MOTORS  225 


SUMMARY  OF  CHAPTER  VIII 

FORCE  ON  WIRE  IN  MAGNETIC  FIELD.  When  a  wire 
carrying  an  electric  current  lies  in  a  magnetic  field  at  right 
angles  to  the  lines,  there  is  a  force  on  the  wire  due  to  the  reac= 
tion  between  the  circular  field  about  the  wire  and  the  field  in 
which  the  wire  lies.  The  amount  of  this  force  can  be  found 
from  the  equation: 

IHZ 

=    "- 


TORQUE.  The  torque  of  the  armature  of  a  motor  is  due 
to  the  above  force,  and  is  computed  from  the  following  equa- 
tion: 


10 

where  I  =  current  in  armature  wire  in  amperes  ; 

H=  field  strength  perpendicular  to  wire  in  gausses; 
1=  total  length  of  active   wire  in  centimeters; 
r  =  radius  of  coil  in  centimeters  ; 
T-=  torque  in  dyne-centimeters. 

MOTOR  EFFECT  IN  GENERATORS.  This  armature  torque 
in  a  motor  is  the  opposing  torque,  which  the  machine  has  to 
overcome  when  running  as  a  generator  with  same  field 
strength  and  delivering  same  current.  This  helps  us  to 
understand  how  power  is  lost  in  producing  eddy  currents. 

The  opposing  current  set  up  by  Eddy  Currents  in  the  bobbin, 
on  which  moving  coils  of  galvanometers,  ammeters,  etc.,  are 
wound,  is  used  to  damp  the  oscillations  of  the  coil. 

Eddy  currents  in  motors  and  generators  have  a  torque 
which  opposes  the  motion  and  consumes  power. 

GENERATOR  EFFECT  IN  MOTORS,  COUNTER  E.M.F. 
The  armature  conductors  of  a  motor  cut  lines  of  force  and 
set  up  an  E.M.F.  opposite  in  direction  to  the  impressed  E.M.F. 

The  current  flowing  through  the  armature  is,  then,  under  a 
pressure  equal  to  the  difference  between  the  Impressed 
Voltage  and  Counter  E.M.F.  The  value  of  the  current  can 
be  found  from  the  equation: 

Ex  -Ec 
A  "•»-»       % 


226  ELEMENTS   OF   ELECTRICITY 

AXIS  OF  COMMUTATION.  BACKWARD  LEAD.  The 
position  of  the  brushes  for  a  motor  for  sparkless  commuta- 
tion is  on  an  axis  back  of  the  neutral  axis,  instead  of  ahead, 
as  in  a  generator. 

DIRECTION  OF  ROTATION.  Extend  the  thumb,  fore- 
finger and  middle  finger  of  the  right  hand,  at  right  angles 
to  one  another,  as  in  the  rule  for  the  generator.  When 
the  middle  finger  points  in  the  direction  of  the  counter  E.M.F., 
the  forefinger  in  direction  of  the  flux,  the  thumb  will  indi- 
cate the  direction  of  necessary  rotation. 

Any  motor  must  be  started  slowly  by  means  of  a  starting 
box,  which  puts  resistance  in  series  with  armature,  to  keep 
the  current  in  the  armature  from  becoming  excessive,  until 
a  counter  E.M.F.  is  set  up  by  the  motion  of  the  armature. 

Motors  are  divided  into  three  general  types. 

(1)  Shunt;   field   shunted   around   the   armature. 
Nearly   constant  speed. 

Low  starting  torque. 

Races  when  field  is  broken,  thus  the  necessity  of  "  no- 
field  "  release. 
Speed  controlled  by  varying  field  strength. 

(2)  SERIES;  field  in  series  with  armature. 
Speed  varies  with  load. 

Large  torque  at  slow  speeds,  thus  large  starting  torque. 
Races  on  "  no-load,"  thus  the  necessity  of  having  load 
permanently  attached. 

(3)  (a)  COMPOUND  (Differential) ;   series  field  coils  buck- 
ing shunt  coils. 

Exaggerated  shunt  characteristics. 

Low  starting  torque. 

Constant  speed  at  all  loads  within  limits. 

(b)  COMPOUND    (Cumulative);   series   field   coils   aiding 
shunt  coils. 

Exaggerated   series   characteristics. 

Will  not  race  at  "  no-load." 

Large  torque  at  low  speeds,  hence  high  starting  torque. 

Speed  varies  greatly  with  load. 

The  characteristics  of  motors  and  generators  of  same  type 
are  very  similar  if  speed  of  motors  is  compared  to  voltage 
of  generators. 


MOTORS 


227 


In 
If 

FIG.  179a. 


PROBLEMS    ON   CHAPTER   VIII. 

17-8.  A  motor  is  running  on  115  volts  and  taking  7.4  amperes 
in  the  armature.  The  armature  resistance  of  the  motor  is  .32 
ohm.  What  counter  E.M.F.  is  being  generated? 

18-8.  From  following  data  compute  the  armature  resistance 
of  motor: 

Impressed   v.oltage  =  220   volts. 
Counter  E.M.F. -214  volts. 
Armature   current  =  12  amperes. 

19-8.  Draw  resultant  field   if   poles  and  current,  Fig.  179a,  are 
as  marked.     Indicate  direction  of  force 
on  each  side  of  loop  and  direction  of 
tendency  of  rotation. 

20-8.  A  and  B,  Fig.  165,  are  each 
14  inches  long  and  carry  20  amperes. 
The  field  intensity  in  which  they  lie  is 
20,000  gausses.  What  is  the  force 
tending  to  make  each  move? 

21-8.  Assume  that  the  distance  from  A  to  B,  Fig.  165,  is  10 
inches.  What  is  the  torque  of  loop? 

22-8.  Suppose  machine  in  Problem  18  were  running  as  a  gen- 
erator at  same  speed,  and  delivering  the  12  amperes  instead  of 
receiving  them;  field  unchanged,  (a)  What  would  the  brush 
potential  of  the  machine  be?  (6)  What  would  be  its  output 
in  K.W.? 

23-8.  There  are  400  conductors  on  a  D.C.  armature.  Assuming 
70  per  cent  of  these  lie  in  a  magnetic  field  of  6000  gausses,  how 
many  pounds  are  acting  on  armature  when  each  conductor  carries 
20  amperes?  Length  of  each  conductor  is  12  inches. 

24-8.  There  are  600  conductors  on  D.C.  armature,  Fig.  1796, 
80  per  cent  of  which  lie  in  the  magnetic  field.  Diameter  of  arma- 
ture is  1 5  inches.  Width  of  pole  face  parallel  to  shaft  is  25  inches. 
Average  intensity  of  field  is  5000  gausses.  Armature  takes  a 
total  current  of  60  amperes.  What  is  force  in  pounds  on  each 
active  conductor? 

25-8.  (a)  What  is  total  force  on  active  conductors  in  Problem 
24?  (6)  What  torque  in  Ib.ft.  is  developed? 

26-8.  (a)  If  speed  of  motor  in  Problem  24  is  1200  R.P.M.  what 
horse  power  is  transmitted  to  pulley,  allowing  90  per  cent  efficiency? 
(6)  What  voltage  must  be  applied  to  motor? 


228 


ELEMENTS  OF   ELECTRICITY 


27-8.  In  Problem  24,  the  width  of  the  pole  face  parallel  to 
shaft  is  25  inches.  Assume  width  in  the  other  direction  to  be 
9.2  inches.  What  is  the  counter  E.M.F.  at  speed  given? 

28-8.  (a)  What  is  the  resistance  through  one  path  in  armature 
of  Problem  24?  (6)  What  is  armature  resistance  from  brush  to 
Brush?  (Use  data  obtained  in  Problems  24,  25  and  26.) 

29-8.  In  a  shunt  motor  the  speed  falls  from  1600  R.P.M.  at 
no  load  to  1480  R.P.M.  at  full  load.  What  is  the  speed  regulation? 


FIG.  1796. 

30-8.  The  armature  of  a  shunt  motor  has  a  resistance  of  2  ohms. 
The  field  has  a  resistance  of  200  ohms.  What  current  will  flow 
in  armature  when  110  volts  are  applied  to  terminals  with  machine 
at  rest?  (6)  Through  fields  wTith  machine  at  rest? 

31-8.  If  motor  in  Problem  30  were  turning  fast  enough  to 
develop  108  volts  counter  E.M.F.,  what  current  would  flow  in 
the  armature  circuit?  (6)  In  the  field  circuit? 

32-8.  Armature,  Fig.  172,  resistance  is  3  ohms.  Field  resist- 
ance =  224  ohms.  Resistance  of  starting  box  is  divided  as  follows : 

1  to  2=2  ohms 

2  to  3  =  2      " 

3  to  4=3     " 

4  to  5  =  2     " 


MOTORS  229 

Voltage  of  line  =  112  volts,  (a)  What  is  starting  current  of  mo- 
tor? (6)  If  counter  E.M.F.  of  motor  is  75  volts  when  running 
on  point  2  of  starting  box,  what  current  is  armature  then  taking? 

33-8.  If  motor  in  Problem  32  takes  3  amperes  when  running  on 
point  4,  what  is  its  counter  E.M.F.  ? 

34-8.  A  shunt  motor  takes  a  total  current  of  80  amperes  from 
115  volt  mains.  The  resistance  of  the  armature  is  .04  ohm.  Resist- 
ance of  the  field  is  60  ohms,  (a)  What  current  does  each  take? 
(6)  What  power  is  used  up  in  heating  the  field  and  armature? 

35-8.  What  current  does  field  take  in  Problem  32  (a)  and  in 
Problem  32  (6)? 

36-8.  (a)  What  is  the  total  power  taken  by  motor  in  Problem 
34?  (6)  Why  is  this  greater  than  the  power  used  in  heating  the 
field  and  armature  as  computed  in  Problem  34?  (c)  Compute 
the  counter  E.M.F.  of  motor  in  Problem  34. 

37-8.  It  requires  1000  ft.  of  No. '30  (B.  &  S.  Gauge)  copper 
wire  for  the  field  coils  of  a  4-pole,  4-brush  shunt  motor.  The 
armature  has  600  ft.  of  No.  25  copper  wire.  If  the  motor  takes 
a  total  current  of  17  amperes  at  112  volts  (a)  What  is  its  counter 
E.M.F.?  (6)  What  power  is  lost  in  heating  armature  and  field? 

38-8.  (a)  What  voltage  would  motor  in  Problem  37  deliver 
if  run  at  the  same  speed,  and  17  amperes  were  flowing  through  the 
armature?  Field  separately  excited  to  same  degree  of  magnet- 
ization, (b)  What  power  would  it  be  delivering  to  the  line? 

39-8.  A  15-K.W.  110-volt  shunt  generator  has  a  speed  of  900 
R.P.M.  at  full  load.  Field  resistance  is  35  ohms;  armature 
resistance  is  .06  ohm.  (a)  What  are  the  armature  and  field 
currents  at  full  load?  (6)  What  is  E.M.F.  of  generator? 

40-8.  If  machine  in  Problem  39  were  run  as  a  110-volt  motor  and 
had  same  current  flowing  through  the  armature,  at  what  speed 
would  it  run?  , 

41-8.  A  2-pole  220- volt  shunt  motor  takes  50  amperes  at  full  load. 
The  field  resistance  is  100  ohms.  Armature  consists  of  200  active 
conductors  and  has  a  resistance  of  .36  ohm.  <£  =  4,000, 000  lines. 
At  what  speed  does  it  run  at  full  load? 

42-8.  If  motor  in  Problem  41  were  a  4-pole  4-brush  machine, 
and  all  other  data  were  the  same,  what  would  be  the  full  load 
speed? 


230  ELEMENTS   OF    ELECTRICITY 

43-8.  If  motor  in  Problem  41  were  a  4-pole  2-brush  machine  and 
all  other  data  were  the  same,  what  would  be  the  full  load  speed? 

44-8.  If  motor  in  Problem  41  requires  8  amperes  at  no  load, 
what  is  the  speed  regulation? 

45-8.  The  armature  of  a  2-pole,  110-volt  shunt  motor  has  180 
active  conductors  and  a  resistance  of  .05  ohrn.  It  takes  6.5 
amperes  to  run  motor  at  no  load  and  100  amperes  at  full  load. 
The  field  has  a  resistance  of  55  ohms.  <£-  3,000,000  lines.  What 
is  the  speed  regulation  of  the  motor? 

46-8.  How  much  resistance  would  have  to  be  placed  in  series 
with  armature  in  Problem  45  in  order  that  the  starting  current 
may  not  exceed  twice  the  full  load  current? 

47-8.  A  550-volt  series  motor  having  4  poles  and  2  brushes, 
requires  180  K.W.  at  full  load.  Field  and  brush  together  have 
resistance  amounting  to  .041  ohm.  There  are  350  active  con- 
ductors on  armature,  which  has  a  resistance  of  .042  ohm.  Speed 
is  600  R.P.M.  What  is  the  flux  per  pole? 

48-8.  A  110-volt  shunt  motor  has  an  armature  resistance  of 
.8  ohm.  The  field  resistance  is  220  ohms.  The  full  load  speed 
is  1200  R.P.M.  and  the  motor  is  taking  10  amperes.  At  what 
speed  must  this  machine  run  as  a  generator  in  order  to  deliver 
10  amperes  at  110  volts? 

49-8.  (a)  What  is  counter  E.M.F.  of  machine  in  Problem  48 
when  running  at  full  load  as  a  motor?  (6)  What  is  E.M.F.  when 
run  as  a  generator  and  delivering  10  amperes  at  110  volts? 

50-8.  The  motor  of  Problem  41  delivers  12  H.P.  at  full  load. 
What  torque  does  it  develop  at  full  load? 


CHAPTER  IX 
FURTHER    APPLICATIONS 

SOLUTION  OF  SOME  OF  THE  MORE  DIFFICULT  PROBLEMS  ENCOUN- 
TERED IN  ELECTRICAL  PRACTICE 

Efficiency  of  Electrical  Machinery  and  Processes — Transmission  of 
Electrical  Power — Efficiency:  Effect  of  Voltage — Relation  of  Size 
of  Conductor  to  Voltage  of  Transmission — Feeders — Three-wire 
System  —  Kirchhoffs  Laws — Current  Distribution  in  Parallel 
Combinations  of  Battery  Cells  or  Generators — Stray  Power  Losses 
— Commercial  Efficiency  of  Generators  and  Motors — Electrical 
Efficiency  of  Generators — Mechanical  Efficiency  of  Motors. 

148.    Advantages   of     Electrical     Transmission.       The 

advantages  of  electrical  transmission  of  power  are  almost  too 
well  known  to  be  mentioned.  The  simplicity  of  the  mechan- 
ism required  for  electrical  transmission  and  the  immense 
distance  that  electricity  can  be  transmitted  are  the  most 
striking  advantages.  A  couple  of  small  stationary  wires 
are  all  that  is  required  to  convey  enormous  quantities  of 
power  in  this  form  from  one  end  of  the  state  to  the  other. 
Mechanical  power  can  be  transmitted  at  best  but  a  few 
yards,  and  then  only  by  such  cumbersome  moving  parts 
as  belts,  ropes,  chains,  etc. 

The  use  of  compressed  air  for  conveying  mechanical 
power  in  small  quantities  from  one  part  of  a  building  to 
another,  is  perhaps  the  nearest  rival  that  electricity  has, 
but  here  too  the  machinery  is  cumbersome. 

The  efficiency  of  electrical  transmission  can  be  made 
remarkably  high,  even  over  great  distances,  if  certain 
conditions  are  complied  with.  It  will  be  instructive  to 

231 


232  ELEMENTS   OF  ELECTRICITY 

find  out  just  what  these  conditions  arc,  and  to  what  extent 
they  affect  the  efficiency. 

149.  Relation  of  Voltage  of  Generator  to  Efficiency 
of  Transmission.  Suppose  it  is  possible  to  generate  3fl 
kilowatts  at  a  certain  place  and  that  it  is  desired  to  use 
the  power  to  run  a  motor,  say  two  miles  distant.  The 
power  must  be  transmitted  in  a  practical  form  and  at  as 
small  a  loss  as  possible. 

In  Fig.  180,  G  represents  the  30  K.W.  generator,  M 
the  motor  at  a  distance  of  about  two  miles  from  G.  Assume 
for  convenience,  that  the  line  wires  have  a  total  resistance 
of  3  ohms.  Let  us  try  the  effect  of  using  various  voltages 


1.5  Ohms 


1.5  Ohms 


FIG.  180.  —  Generator  delivers  30  K.W.  to  line. 

ranging  from  100  to  10,000  volts,  at  which  we  might  gen- 

erate the  30  K.W.  and  compute  the  corresponding  efficiencies. 

Consider  the  efficiency  of  the  generator  to  be  100  per  cent. 

First,  assume  that  we  generate  at  100  volts.     To  supply 

30  000 

30  K.W.  at  this  voltage,  the  generator  must  deliver  —  J-^~ 

100 

or  300  amperes  to  the  line. 

Now,  to  force  300  amperes  through  a  line  wire  having 
a  resistance  of  3  ohms,  would  require  300X3  or  900  volts; 
which  is  800  more  volts  than  we  are  even  generating.  It 
is  clearly  impossible  to  transmit  300  amperes  through  the 
line  at  this  pressure. 

Let  us  then  assume  that  we  generate  at  200  volts.  To 
supply  30  K.W.  at  this  voltage,  the  generator  must  deliver 

30  000 
only      '        or  150  amperes  to  the  line. 


Here  again,  to  transmit  as  much  as  150  amperes  through 


F  UR  TH  ER  A  P  PLICA  TlONS 


233 


3  ohms,  requires  3X150  or  450  volts,  which  is  250  volts 
more  than  we  assumed  generated. 

30  000 

If  we  generate  at  300  volts,   only      '        or  100  amperes 

oUU 

must  be  transmitted.  This  would  require  100X3  or  300 
volts  "to  force  it  through  the  line  alone.  Since  we  are  gen- 
erating but  300  volts,  this  leaves  no  voltage  to  force  the 
current  through  the  motor  M.  Efficiency  of  transmission, 
then,  is  zero,  since  all  the  power  is  lost  in  the  line. 

If  we  generate  at  400  volts,  it  is  necessary  to  force  but 

OQ  OOf) 

•   '        or  75  amperes,  through  the  line.     The  volts  required 

for  this  would  be  75X3  or  225,  leaving  400-225  or  175 
volts  to  run  the  motor. 

Watts  lost  in  line  =752X3  =  16,875  watts. 

Watts  left  for  motor  =30,000 -16,875  =  13, 125  watts. 

13  125 

Efficiency  of  transmission  =  ^7^  =43.8  per  cent. 

oU,Uu(J 

The  results  of  similar  computations  for  a  range  of  volt- 
ages may  be  tabulated  as  follows: 

30  K.W.  GENERATED  AT  E  VOLTS 


Volts 
E 

Amperes. 
I 

Line  Loss 
in  Volts. 
R  =  3  ohms 

IR 

Line  Loss 
in  Watts. 

72/2 

Volts  Left 
for  Motor 

V 

Watts 
Trans- 
mitted to 
Motor. 
IV 

Efficiency. 
Per  cent. 

100 

300 

900 

Impossible 

200 

150 

450 

it 

300 

100 

300 

30,000 

0 

0 

0 

400 

75 

225 

16,875 

175 

13,125 

43.8 

500 

60 

180 

10,800 

320 

19,200 

63.3 

600 

50 

150 

7,500 

450 

22,500 

75. 

800 

37.5 

112.5 

4,220 

687.5 

25,780 

86 

1,000 

30 

90 

2,700 

910 

27,300 

91 

1,200 

25 

75 

1,875 

1,125 

28,125 

93.8 

1,500 

20 

60 

1,200 

1,440 

28,800 

96 

2,090 

15 

45 

675 

1,955 

29,325 

97.8 

3,000 

10 

30 

300 

2,790 

29,700 

99 

5,000 

6 

IS 

108 

4,994 

29,964 

99.8 

10,000 

3 

9 

27 

9,991 

29,973 

99.9 

234 


ELEMENTS  OF  ELECTRICITY 


The  curve  plotted  in  Fig.  181  shows  the  relation  of 
voltage  to  efficiency  of  transmission,  as  brought  out  by 
the  above  problem.  Note  that  the  efficiency  increases  very 
rapidly  with  the  voltage  until  about  1200  volts  are  reached. 
From  there  on,  the  increase  is  much  slower,  though  the 
efficiency  continues  to  rise  a  little  with  each  increase  of 
voltage.  If  the  voltage  were  the  only  condition  affecting 
the  efficiency,  it  would  be  advisable  to  use  indefinitely 
high  voltages.  The  difficulties  in  the  insulation  of  the 
machine  and  line,  limit  the  voltage  at  the  present  time  to 


400      800     1200    1600    2000    2400    2800   3200 

Voltage  at  Generator 
FIG.  181. — Relation  of  efficiency  of  transmission  to  brush  potential  of  generator. 

some  more  moderate  value.  Yet  the  fact  is  evident  that  as 
far  as  the  line  loss  (I2R\  is  concerned,  the  higher  the  voltage 
used,  the  higher  the  efficiency  of  transmission. 

Problem  1-9.  Construct  a  table  as  above  for  80  K.W.  genera- 
tor and  2  ohm  line,  starting  at  100  volts  and  gradually  increasing 
it  to  10,000  volts.  Plot  curve  between  volts  and  efficiency  of 
transmission. 

Problem  2-9.  Repeat  Problem  1-9,  using  8  K.W.  generator. 

150.  Relation  of  Line  Loss  to  Voltage  of  Generator 
Maintaining  Constant  Output.  On  inspection  of  above 
table,  it  will  become  evident  that  the  line  loss  in  watts 
varies  inversely  as  the  square  of  the  voltage  of  the  generator 
(assuming  a  generator  of  constant  watt  output.) 


FURTHER  APPLICATIONS  235 

For  instance,  the  line  loss  at  500  volts  is  10,800  watts, 

10  800 
while  at   1000  volts  it  is  only       ^      or  2700  watts.      By 

doubling  the  voltage  we  have  quartered  the  line  loss. 

Problem  3-9.  (a)  If  the  line  loss  on  a  110- volt  system  is  4  K.W., 
what  will  it  become  if  the  voltage  of  the  system  is  changed  to  220 
volts?  (6)  To  550  volts?  Assume  same  power  delivered  to  the 
line. 

151.  Relation  of  Size  of  Conductor  to  Voltage  (Line 
Loss  Constant).  In  the  above  paragraphs  we  have  been 
considering  a  means  of  decreasing  the  operating  expenses 
by  decreasing  the  power  lost  in  the  line.  We  have  seen 
that  the  power  lost  decreases  as  the  square  of  the  voltage 
of  transmission.  It  is  also  important  to  consider  the 
effect  on  the  original  cost  of  transmission  if  we  raise  the 
voltage.  In  analyzing  this  point,  we  will  assume  that  we 
have  fixed  upon  an  amount  of  power  which  we  are  willing 
shall  be  lost  in  the  line,  and  that  we  keep  this  constant. 

By  referring  again  to  the  above  table,  we  find  that  we 
have  a  loss  of  1.2  K.W.,  when  the  generator  voltage  was 
1500  volts.  Let  us  assume  this  to  be  the  line  loss  we  are 
willing  shall  be  kept  constant  for  economical  reasons.  By 
doubling  the  voltage  to  3000  volts,  we  of  course  decreased 

1.2 

the  line  loss  to  -~  or  .3  K.W.     But  since  we  are  allowed 

a  line  loss  of  1.2  K.W.,  we  may  now  increase  our  line  re- 
sistance by  using  a  wire  of  a  diameter  enough  smaller  to 
make  up  this  line  loss. 

The  resistance  of  the  line  has  been  kept  at  3  ohms.  At 
1500  volts,  the  line  current  was  20  amperes;  line  loss  = 
202X3=  1.200  K.W. 

At  3,000  volts,  line  current  was  10  amperes;  line  loss  = 
102X3=.300  K.W. 

But  being  allowed  1.2  K.W.  line  loss,  if  we  use  the  3000 
volts,  and  10  amperes  line  current,  we  may  use  a  wire  of 
12  ohms,  instead  of  3  ohms.  Then  the  line  loss  would 


236  ELEMENTS  OF  ELECTRICITY 

become  102X12  =  1.2  K.W.,  the  same  value  it  had  at  1500 
volts  with  a  3  ohm  line.  That  is,  by  doubling  the  voltage 
we  are  able  to  use  a  line  wire  of  four  times  the  resistance, 
and  not  increase  the  line  loss. 

This  may  be  stated  in  more  general  terms  as  follows: 

The  line  resistance  for  a  constant  line  loss  varies  directly 
as  the  square  of  the  voltage  of  transmission. 

Since  the  resistance  of  a  wire  varies  directly  as  the  length 
and  inversely  as  the  cross-section  area,  the  above  law  may 
be  applied  as  meaning,  that  with  the  same  line,  loss  we 
may  either: 

(a)  Use  the  same  size  wire  and  transmit  four  times  as 
far  at  double  voltage,  nine  times  as  far  at  triple  voltage, 
etc.,  or 

(6)  Transmit  the  same  distance  and  use  wire  of  one- 
quarter  cross-section  or  of  one-quarter  weight  if  we  double 
the  voltage;  one-ninth  weight,  if  we  triple  the  voltage,  etc. 

It  can  be  seen  from  (6)  that  in  transmitting  electric 
power  between  any  two  points  at  given  line  loss,  a  great 
saving  in  the  cost  of  copper  can  be  made  by  transmitting 
at  a  high  voltage.  In  fact,  since  the  cost  of  installed  mod- 
erate-sized copper  wire  is  almost  directly  proportional  to 
its  weight,  we  may  say  that  as  far  as  the  cost  of  the  line 
wire  is  concerned,  the  initial  expense  varies  inversely  with 
the  square  of  the  voltage  of  transmission. 

Problem  4-9.  A  110-volt  line  is  to  transmit  the  same  power  at 
the  same  line  loss  as  a  220-volt  line  of  equal  length.  How  will 
the  size  of  the  line  wires  of  the  two  systems  compare? 

Problem  5-9.  If  No.  00  (B.  &  S.)  wire  is  used  for  the  110-volt 
line  of  Proolem  4-9,  what  size  wire  would  be  necessary  for  the 
220  volt  line? 

In  choosing  the  line  wire  to  be  used  in  the  installation  of 
a  system,  such  a  size  must  be  used  as  will  maintain  a  £air 
balance  between  the  initial  cost  and  the  cost  of  running, 
i.e.,  the  line  loss.  The  initial  cost  increases  as  the  size 


FURTHER  APPLICATIONS 


237 


of  the  wire  increases,  while  the  cost  of  running  (line  loss) 
decreases  as  the  size  increases.1 

152.  Feeders.  In  order  to  avoid  a  heavy  line  loss  in 
both  power  and  voltage,  it  is  customary  in  many  low  volt- 
age systems,  i.e.,  550  volts  or  under,  to  parallel  the  main 
conductor  with  a  second  conductor  called  a  FEEDER. 

Such  a  feeder  is  often  run  alongside  of  a  trolley  wire  and 
joined  to  it  at  regular  intervals. 

This  arrangement  is  merely  a  case  of  parallel  conductors, 
but  it  presents  some  interesting  problems  in  current,  volt- 
age, and  power  distribution. 


Track  Resistance  =  .04  Ohms  per  Mile 
FIG.  182. — Trolley  line  with  feeder.     Cars  at  juncture  of  feeder  and  trolley  wire. 

Example.     Trolley  in  Fig.  182  is   No.  0,  hard-drawn  copper, 
.520  ohm  per  mile.      Feeder  is  No.  0000  annealed  copper  .258 
ohm  per  mile.     Track  resistance  is  .04  ohm  per  mile. 
Car  No.  I  is  1  mile  from  Generator  station. 
"        II  is  2  miles  from  Car  I. 
"      III  is  2  miles  from  Car  II. 

Feeder  extends  4  miles  from  Station  and  is  tied  to  trolley  every 
half  mile. 
Find: 

(a)  Volts  lost  between  Generator  and     I. 

(I  1C.  (I  T  «  TT 

"  "  "  II  "        III. 

(b)  Voltage  across  each  car. 

(c)  Power  lost  in  each  section  of  line. 

(rf)  Efficiency  of  transmission  to  three  cars. 

1  Lord  Kelvin  deduced  the  following  rule  with  regard  to  the  most 
economical  area  of  conductor  at  given  voltage  and  distance: 

The  conductor  should  be  of  such  an  area  that  the  value  of  power 
lost  per  year  in  the  line,  equals  the  interest  per  year  on  the  money 
invested  in  the  conductor. 


238  ELEMENTS  OF   ELECTRICITY 

Solution.  The  feeder  and  trolley  may  be  considered  as  two 
parallel  circuits  with  a  combined  resistance  per  mile,  found  as 
follows : 

Resistance  Conductivity 

per  mile  per  mile 

.530  ohm  1.89  mhos 

.258  ohm  3.88  mhos 


5.77  mhos  =  — — ,  or  .173  ohm. 
o.77 


If  the  trolley  and  feeder  had  both  been  of  the  same  kind  of 
copper,  it  would  have  been  possible  to  have  considered  them  as 
one  wire,  the  cross-section  area  of  which  was  equal  to  the  sum 
of  the  cross-section  areas  of  each.  Then  the  resistance  per  mile 

TfJ 

could  have  been  found  from  the  equation  R=~7  as  explained  in 

Chapter  V. 

The  trolley  resistance  between  Gen.  and  Car   I 

=       IX.  173  =.173  ohm. 
Car    I  and  Car  II 

=       2X.  173=.  346     " 
Car  II  and  Car  III 

=  .173X.330=.703     " 
"     total  "  "         Gen.  and  Car      I 

=  .173+    .04=. 214     " 
Car    I  and  Car  II 

=  .346+    .08=. 426     " 
Car  II  and  Car  III 

=  .703+    .08=.  783     " 
Current  between  Car     II  and  Car  III  =  60  amps. 

I  and     "     .11  =  60  +  30         =  90     " 
"       Gen.       and     "  1  =  90  +  40         =130     " 

Drop  in  line  from  Gen.  to  Car  I 

=  130  amps. X. 214  ohms  =  27.8  volts. 
"     across  Car  I  =560  -27.8  =  532  volts. 

"     in  line  from  Car  I  to  Car  II 

=  90  amps. X. 426  ohms  =  38.3  volts. 
"     across  Car  II  =  532  - 38.3  =  494  volts. 

"      in  line  from  Car  II  to  Car  III 

=  60 X. 783  =  47  volts. 
"     across  Car  III  -494-47  =447  volts. 


FURTHER  APPLICATIONS 


239 


Power  lost  between  Gen.  and  Car      1  =  ISO2 X. 214  =  3620   watts. 
Car  I  and  Car    11=  902X.426  =  3450       " 
Car  II and  Car  111=   602X. 783 -2820      " 

Total  power  lost  in  line  =9890       " 

Power  delivered  by  Gen.  =  560 X 130  =  72,800  watts. 

to   Cars  =72,800  -  98,900  =62,900       " 
or 


Power  delivered  to  Car 


I  =  532  X  40  =  21 ,280  watts. 
11  =  494X30  =  14,820      " 
111  =  447X60  =  26,820      " 


to  Cars  =62,920     " 

62  900 
Efficiency  of  transmission  =  — j— —  =  86.4  per  cent. 


n/:268Ohm  perMUe 
r  T* ——i      E          .530  Ohm 


FIG.  183. — Trolley  line  with  feeder  cars  not  at  juncture  of  trolley  and  feeder. 

When  the  cars  do  not  happen  to  be  exactly  at  a  junction  of 
the  feeder  and  trolley,  the  problem  becomes  a  little  more  involved. 

Assume,  in  Fig  183,  the  hard-drawn  trolley  wire  .530  ohm  per 
mile  and  feeder  .258  ohm  per  mile,  joined  every  mile  to  the  trolley. 

Distance  between  Gen.  and  Car  1  =  1  mile. 

Car  I  and  Car  II  =  H  miles. 
Car  II  and  Car  III  =  li-  miles. 

Track  resistance  =  .04  ohm  per  mile. 
Solution : 

As  before  the  combined  resistance  per  mile  of  trolley  =  .173  ohm. 

As  before,  the  current  through  section  .4  #  =  130  amperes. 
Volts  lost  in  line  between  Gen.  and  Car  I  =  130  *  (.1 73  +  .04)  =  27.S. 
Drop  across  Car  1  =  560-27.8  =  532  volts. 

It  is  necessary  now  to  investigate  how  the  current  divides  among 
the  sections  CD,  HF,  and  FJ. 

Let  the  current  through  CD  —  x  amperes. 

Then  the  current  in  FJ,  plus  the  current  in  CD  must  equal  60 
amperes,  since  these  two  branches  both  feed  to  Car  III. 


240  ELEMENTS  QF  ELECTRICITY 

Thus  we  may  write  the  equation, 

Current  in  FJ  =  (60—  x)  amperes. 

The  current   in  HF  must  be  30  amperes  more  than  that  in  FJ, 
since  HF  feeds  30  amperes  to  Car  ]j  before  it  feeds  to  FJ. 
Hence : 

Current  in  /7^  =  60-a:  +  30=(90-z)  amperes. 

Resistance  of  CD  =  .258  ohm. 

Drop  along  CD      =  .258s  volt. 

, . 530 

Resistance  of  HF  =  resistance  FJ  =  ^—  =  .265  ohm. 

Drop  along  HF     =  .265  X  (90  -x)  =  (23.85  -  .265s)  volts. 
Drop  along  IV      =  .265X(60-z|=(15.9-.265x)_™lts. 
Drop  along  HJ      =drop  along  HF  +  drop  along  FJ. 

=  (23.85  -.265z)  +  (15.9  -.265z) 

=  39.75 -.53s  volts. 

But  since  CD  and  ///  are  in  parallel,  the  drop  along  each  must 
be  the  same  (assuming  ties  have  no  resistance). 
Hence : 

39.75-.53s=.258s 
.788a*  =  39. 75 

39.75     _ 
x  =  =  50 . 4  amperes. 

.788 

Current  through  CD  —  50 . 4  amperes. 

#£•  =  90-50.4  =  39.6      " 
"         ^^  =  60-50. 4=  9.6      " 
Trolley  drop  between  Car  I  and  II 

=drop  along  J5CY+drop  along  HF 

530 

=  (90X.173)  +  (39.6X: )=26  volts. 

Total  drop  between  Car  I  and  II 

=  26  +  (90X1.5X.04)=31.4volts. 
Trolley  drop  between  Car  II  and  III 

=  drop    along   /*V  +  drop   along   JE. 

530 
=  (9.6X^-— )  + (60 X. 530)  =35.2    volts. 


FURTHER  APPLICATIONS  241 

Total  drop  between  Car  II  and  III 

=  35.2  +  (60X1.5X.04)=38.8    volts. 

Power  lost  in  line  and  efficiency  of  transmission  may  now  be 
found  as  in  previous  example. 

Problem  6-9.  An  eight-mile  electric  railway  has  a  No.  0  trolley 
wire  of  hard-drawn  copper.  A  feeder,  No.  0000  soft  copper  wire, 
extends  from  the  station  5  miles  along  the  trolley  wire  and  is  tied 
to  it  every  mile.  At  a  certain  instant  there  are  4  cars  on  the  line. 
Car  I  is  2  miles  from  station,  taking  40  amperes,  Car  II  is  3£  miles 
from  station,  taking  60  amperes.  Car  III  is  6  miles  from  station, 
taking  30  amperes.  Car  IV  is  7?  miles  from  station,  taking  50 
amperes.  Track  resistance  is  .05  ohm  per  mile.  What  must  terminal 
voltage  of  generator  be  in  order  that  Car  IV  may  have  a  pressure 
of  500  volts? 

Problem  7-9.  If  the  resistance  of  Generator  in  Problem  6  is  .08 
ohm,  what  E.M.F.  must  it  generate? 

153.  Three-Wire  System.  We  have  seen  that  the 
amount  t>f  wire  needed  for  distribution  with  given  line 
loss  varies  inversely  as  the  square  of  the  voltage  of  trans- 
mission. 

Whatever  may  be  the  voltage  of  transmission,  the  usual 
voltage  to  be  delivered  to  the  customer  is  from  110  to  115 
volts.  In  order  to  gain  the  advantage  of  transmitting  a 
direct  current  at  a  high  voltage,  and  still  deliver  power 
at  but  110  volts,  an  ingenious 

system    has    been    developed   g 

called  the  Three-wire  System. 

Fig.    184   is  a  diagram    of 
this    scheme.      The    110-volt 
generators    G\    and    G2    are    ' 
joined  in  series.     Three  wires 
A,  B  and  C  are  run  as  shown, 
C   being    called    the    neutral    v 
wire.     The  voltage  between  A     FlG'  1M'lS'ES5JSFem  using 
and  B  will  be  220  volts.     If 
we  cause  all  the  power  to  be  distributed  on  the  two  wires 


242 


ELEMENTS   OF  ELECTRICITY 


A  and  B,  the  current  flowing  out  along  A  and  back  through 
B}  we  will  have  to  use  wires  but  one-quarter  as  large  as 
though  we  were  transmitting  the  same  power  at  110-volts, 
allowing  same  line  loss. 

The  system  is  generally  so  "  balanced  "  that  this  is 
approximately  the  case.  To  "  balance  "  a  three-wire  sys- 
tem, it  is  necessary  to  so  arrange  the  loads  on  each  side  of 
the  neutral  that  the  neutral  carries  practically  no  current. 

Fig.  185  shows  a  balanced  three-wire  system.  Notice  that 
if  the  lamps  are  all  the  same  size,  no  current  is  taken  from 


54321  B 

FIG.  185. — Balanced  three-wire  system. 

the  generators  or  returned  to  them  by  the  neutral.  The 
effect  is  the  same  as  running  the  lamps  in  sets  of  two  in 
series  across  220  volts. 

Fig.  186  shows  an  unbalanced  three-wire  system.  The 
figures  represent  the  proportional  parts  of  the  current 
carried  by  each  section  of  the  lead  wires. 

Since  B  is  returning  less  current  than  A  is  supplying, 
the  neutral  must  help  B  return  the  current  from  the  lamps. 
If  B  were  returning  more  current  than  A  was  supplying 
to  the  lamps,  then  the  neutral  would  help  A  deliver  current 
to  the  lamps.  Although  it  is  intended  that  the  system 
be  kept  balanced,  the  neutral  is  generally  made  of  the  same 
size  as  A  and  B,  As  each  wire  is  one-quarter  the  size  of 


FURTHER  APPLICATIONS 


243 


each  wire  in  a  two-wire  system  of  110  volts,  only  three- 
eighths  as  much  copper  is  used  in  a  three-wire  as  in  a  two- 
wire  system  in  transmitting  equal  power. 


Neutral 


43  B 

FIG.  186. — Unbalanced  three-wire  system. 

It  is  costly  to  install  and  operate  two  generators,  so  a 
special  three-wire  generator  has  been  invented  which  allows 
this  system  to  be  operated  by  one  machine.  See  advanced 
text-books  for  a  description  of  this  generator. 

Example.     Assume  that  each  lamp  in  Fig.  187  takes  1  ampere. 
M 


.2  Ohms 


.2  Ohms 


.3  Ohms 


.2  Ohms 


N  .2  Ohms  E  .3  Ohms  F 

FIG.  187. — Balanced  three-wire  system. 


Find: 

(1)  Line  drop. 

(2)  Volts  across  each  set  of  lamps. 

(3)  Line  loss. 


244 


ELEMENTS   OF  ELECTRICITY 


Since  the  system  is  balanced,  no  current  flows  in  neutral.     The 
system,  then,  amounts  to  a  two-wire  220-volt  line. 
Current  from  A  to  B  =  5  amperes. 

Current  from  F  to  E  =  5  amperes. 

Line  resistance  from    A  to  E=.3  +  .3  =  .6  ohm. 
Line  drop  in  section        AB  and  FE  =  5X.6  =  3.0  volts. 
Current  from  M  to  A  =7  amperes. 

E  to  N  =  7  amperes. 
MA"  and  EN  =  A  ohm. 

=  .4X7  =  2.8  volts. 


Resistance  of  line 
Line  drop 


Drop  across  lamps  A  to  #  =  220-2.8  =  217.2  volts. 

"  each  set  of  lamps  AC  and  CE=  108.6  volts. 

"     BDandBF 

_  (217.2 -3) 

2 

Line  loss  MA  and  EN  =  72  X  .4  =  19.6  watts. 
=52X.6  =  15      watts. 


=  34.6   watts. 


=  107.1. 


Total  line  loss 
M 


.3  Ohms 


.3  Ohms 


N     .2  Ohms    E  .3  Ohms 

FIG.  188. — Unbalanced  three-wire  system. 

Example.     Assume  each  lamp  in  Fig.  188  takes  1  ampere. 
Find: 

(1)  Line  drop. 

(2)  Line  loss. 

(3)  Voltage  across  each  set  of  lamps. 

(4)  Efficiency  of  transmission. 

Since  the  system  is  unbalanced,  there  will  be  a  current  flowing 
in  the  neutral. 


FURTHER  APPLICATIONS  245 


The  current  in  AB  =  5  amperes 

"  DF=2  " 

CD  =  3  " 
AC  =  2 

MA  =7  " 

CE  =  1  " 
0(7  =  4       " 


Drop  along  AfA=7X.2  ohm  =  1.4 volts. 
A£  =  5X.3      "    =1.5      " 

((  T\fy O  \y    o          ii       f\  >. 

u  Of?  =  4-  V  9       **     R          ^ 

EF  =  2X.3      "    =.6 
"  NE=3X.2      "    =.6        " 

Voltage  across  AC  =  1 10  -  (1 .4  +  .8)     =  107 . 8  volts. 
A#  =  220-(1.4  +  .6)     =218 
(7^  =  218 -107.8  =110.2     " 

J5D  =  107.8-(1.5  +  .9)  =105.4      " 
BF =218 -(1.5 +  .6)     =215.9      " 
7)^=215.9-105.4        =110.5      u 
Line  loss  in  MA  =  72X.2  =  9.8   watts. 


Total  line  loss  =  26.2  watts. 

Power  delivered  by6?1  =  110X7         =   770  watts. 
G2  =  110X3         =   330      " 

Total  power  delivered  =1100      " 

Power  used  by  lamps  =  1100  -26.2    =  1073.8  watts. 

VK  •  t  .  •    •         1073.8 

l^mciency  ot  transmission  =—  -  =  97.o  per  cent. 

1  1  00 

Problem  8-9.     Assume   each          ft 
lamp  in  Fig.  189  takes  4  amperes. 

Find: 


=  22X.  3  =  1.2 


(a)  Current  in  A  B,  CD,  EF.         S 
C6)  Line  drop  in  A  B.  CD,  EF.     -| ^ 

i       7  A  Ohms 

(c)  Line  loss. 


(d)  Voltage  across  BD  and  DE.    >  {  Q2  \  66666666 

/   \  T^OI    •  PI  •      •  Q     \       S 

(e)  Erhciency  of  transmission,      a 
(/)  Resistance    of    each   set    of 


lamps.  FIG.  189.—  Unbalanced  three-wire  system. 


246  ELEMENTS  OF  ELECTRICITY 

Problem  9-9.  Suppose  neutral  in  Problem  8  broke  at  x,  what 
voltage  would  there  be  across  BD  and  DE?  (Assume  resistance 
of  lamps  constant). 

154.  More  Difficult  Applications  of  Ohm's  Law. 
Kirchhoff 's  Laws.  Several  further  laws  have  been  deduced 
from-  Ohm's  Law  and  called  Kirchhoff's  Laws,  in  honor 
of  the  man  who  first  stated  them.  They  are  merely  an 
extension  of  Ohm's  Law  and  need  not  be  learned  separately. 
In  fact,  we  unconsciously  have  been  applying  them  to 
most  of  the  problems  dealing  with  the  relation  of  cur- 
rent, resistance  and  voltage,  in  the  more  complicated 
systems. 

It  is  well,  however,  to  make  a  definite  statement  of  the 
principles  under  the  proper  heading  of  KIRCHHOFF'S  LAWS, 
in  order  to  apply  them  more  directly  to  certain  complex 
problems  in  distribution. 

KirchhofT  stated: 

First  Law.  That  at  any  point  in  a  circuit,  there  is  as 
much  current  flowing  away  from  the  point  as  there  is  flowing 
to  it. 

Second  Law.  That  the  sum  of  the  several  IR  drops 
around  any  one  path  of  an  electric  circuit  equals  the  sum 
of  the  E.M.F.'s  impressed  on  that  same  path.  Care  must 
be  taken  to  get  the  algebraic  signs  of  the  E.M.F.'s  correct. 
If  there  is  no  source  of  E.M.F.  in  any  given  circuit,  then 
the  sum  of  the  IR  drops  in  one  direction  equals  the  sum 
of  IR  drops  in  the  other  direction. 

First  Law.  This  can  best  be  explained  by  referring  to 
diagram  in  Fig.  190.  Assume  each  lamp  to  take  1  ampere 
and  consider  the  point  B. 

Since  there  are  8  amperes  flowing  away  from  B,  there 
must  be  8  amperes  flowing  to  B  along  AB. 

Consider  point  E. 

Since  there  are  8  amperes  flowing  to  point  E,  there  must 
be  8  amperes  flowing  away  from  E. 

We  have  unconsciously  used  this  law  again  and  again, 


FURTHER  APPLICATIONS.  247 

considering  that  it  was  too  obvious,  from  the  very  nature 
of  an  electric  current,  to  need  demonstration. 

Second  Law.  This  law  is  as  obvious  as  the  first,  but 
more  difficult  to  state  clearly.  Referring  again  to  Fig.  190, 
the  E.M.F.  of  the  generator  G  must  equal  the  IR  drop  in 
AB+BE+EF+ihe  IR  drop  of  the  generator  itself. 

Or  considering  another  complete  circuit  on  the  diagram; 
the  E.M.F.  of  generator  must  equal  the  IR  drop  in  AB  + 
BC  +  CD+ED+FE+gQneT&toY  IR  drop.  Or  again,  con- 
sidering the  circuit  BCDE,  which  contains  no  source  of 
E.M.F.,  we  can  say  that  the  IR  drop  of  BE  must  equal 
the  IR  drop  of  BC+CD+DE,  since  the  IR  drop  of  BE 


is  counter  clockwise  in  the  circuit  BCDE  and  the  IR  drop 
of  BC+CD+DE  is  clockwise. 

We  have  been  accustomed  to  use  this  same  principle 
under  the  two  separate  rules  (1st)  that  the  IR  drop  around 
any  path  in  an  electric  circuit  equals  the  sum  of  the  impressed 
E.M.F.'s,  and  (2d)  that  the  voltage  drop  along  parallel 
paths  between  two  points  is  the  same. 

Thus  we  would  have  said  that  the  voltage  across  BE  equals 
the  voltage  across  BC+CD+DE  because  both  the  path 
BE  and  the  path  BC+CD+DE  are  in  parallel  between 
the  points  B  and  E. 

There  are,  therefore,  no  new  facts  to  be  learned  from 
Kirchhoff's  laws,  but  slightly  different  viewpoints  from  which 
to  regard  familiar  facts.  The  following  example  will  make, 
this  clear: 


248  ELEMENTS  OF  ELECTRICITY 

Example.  Assume  trolley  line,  Fig.  191,  to  be  fed  by  two 
generators,  G,  of  560  volts,  and  G2  of  555  volts.  Two  cars  are 
on  the  line;  Car  I  taking  300  amperes,  Car  II,  200  amperes.  The 
resistance  of  trolley  and  track  as  marked. 

Find: 

(1)  Voltage  across  each  car. 

(2)  Efficiency  of  transmission. 

Solution.  Assume  current  to  flow  in  direction  indicated  in 
different  sections. 

NOTE. — If  on  solution  any  current  values  come  out  negative, 
it  merely  means,  as  in  solving  force  diagrams  in  mechanics,  that  the 
direction  of  the  current  should  be  the  reverse  of  that  indicated; 
the  numerical  result  will  be  correct. 

Let  x  =  current  in  section  AB. 

Then  x  =  current  in  section  HK. 

"     x  -300=       "  "        BCandKF. 

500  -x=       "  CDandEF. 

A . B C D 

.39  Ohm         /f  .13  Ohm 

11-200  Amps. 


.04  Ohm  K  .06  Ohm  F  .02  Ohm       E 

FIG.  191. — Trolley  line  with  generators  at  ends  of  line. 

The  above  assumptions  are  made  in  accordance  with  the  common- 
sense  law  that  the  amount  of  current  which  flows  away  from  a 
point  must  be  the  same  as  that  which  flows  to  it.  This,  as  we 
have  seen,  is  called  Kirchhoff's  "  First  Law." 

Voltage  across  Car  II  equals 

(1)  555 -.15(500 -a?). 
Voltage  across  Car  II  also  equals 

(2)  560 -.30z-. 45  Or-- 300). 
Therefore 

(3)  535 -.15(500 -x}  =560 -30z -A5(x -300). 

These  assumptions  are  made  in  accordance  with  the  principle 
stated  in  Chapter  III,  that  the  voltage  across  a  parallel  combina- 
tion is  the  same  as  that  across  any  and  all  of  the  parallel  paths 
between  the  same  two  points.  This,  as  we  have  seen,  is  part  of 
"  Kirchhoff's  Second  Law."  Thus  the  voltage  across  Car  II  must 
be  the  same  whether  computed  from  the  voltage  of  G2  minus  line 


FURTHER  APPLICATIONS 


249 


drop,  or  from  the  voltage  of  Gl  minus  the  line  drop  from  that 
Generator. 

Solving   (3)   for  x, 

x  =  239  amperes     =  current  in  A  B  and  HK. 

Voltage  across  Car  I  =  560  - 239  X  .30  =  488.3  volts. 

Current   in   CD         -500-239          =261  amperes. 

Voltage  across  Car  II  - 555  -261  X  .15  =  515.8  volts. 

Current  in  BC  =  239  —  300  =  —61  amperes  (the  minus  sign  means 
it  must  be  flowing  in  the  direction  opposite  to  that  marked). 

Power  delivered  by  Gei^  =  560  X  239  =  134,000  watts. 
Gen  2  =  555X261  =  145,000      " 

Total  power  =  279.000  K.W. 

Power  used  by  Car  I  =  488.3  X  300    =  146,800  watts. 
"     11  =  515.8X200    =103,200       " 


Total  power  used  by  cars 


250.000  K.W. 


250 


Efficiency  of  transmission  =  ——=89.7  per  cent. 

2 1 9 

By  means  of  the  principles  illustrated  above,  all  problems 
involving  current  and  voltage  relations  in  "  mesh-work  "  may 
be  solved. 


atil: 


Nl  — 


II 


FIG.  192. 


Problem  10-9.  In  Fig.  192,  each  lamp  takes  12  amperes. 
Resistance  of  AB  =  BC  =  M  ohm.  Resistance  of  CD  =  MK  =  .Q3 
ohm.  Resistance  of  EF  =  KF  =  .Q2  ohm.  What  is  the  voltage 
across  Group  I  and  Group  II,  if  terminal  voltage  of  (71  =  120  volts 
and  of  G2  =  125  volts. 

Problem  11-9.  What  is  the  efficiency  of  transmission  of  system 
in  Problem  10. 


154.  Parallel    Combinations    of    Unlike    Generators    or 
Battery  Cells.      When  battery  cells,  of  either  PRIMARY  or 


250 


ELEMENTS  OF  ELECTRICITY 


STORAGE  type,  are  used  in  series  and  parallel  combination, 
they  are  considered  to  have  the  same  internal  resistance 
and  E.M.F.  This  is  approximately  true,  and  the  process 
of  finding  resulting  voltage  and  current  is  a  matter  of  simple 
addition.  Even  if  the  cells  are  unlike  and  are  joined  in 
series,  the  E.M.F.'s  and  resistances  merely  add  together. 
This  case  need  not  be  considered. 

But  it  is  interesting  to  see  what  will  happen  if  we  have 
two  or  more  cells  of  unlike  E.M.F.  and  resistance,  joined 
in  parallel  and  feeding  a  line.  Such  combinations  represent 
actual  conditions  in  many  telegraph  and  telephone  circuits, 
and  afford  excellent  practice  in  the  application  of  Ohm's 
Law  as  extended  by  Kirchhoff. 

Example.  Consider  Fig.  193.  Cell  El  has  2  volts  E.M.F.  and 
.5  ohm  internal  resistance.  Cell  E2  has  1.4  volts  E.M.F.  and  .8 

ohm  internal  resistance.  When 
they  are  joined  in  parallel  to 
feed  a  line  of  1.7  ohms  resistance: 

Find: 

(a)  Amount  and  direction  of 
current  through  each  cell  and 
through  the  line. 

(6)  Terminal  voltage  (across 
AB). 

Solution.  Assume  current  flows 
as  marked.  There  is  probably 
a  reversed  current  through  E2 
on  account  of  the  higher  E.M.F. 
of  cell  EI.  We  will  assume  this, 
and  if  the  current  value  comes 
out  a  negative  quantity,  we  have 


x-y 


FIG.  193.— Unlike  cells  in  parallel 
feeding  line. 


merely  to  reverse  the  arrow  head. 
Let          x  =  current  through  Elt 
Let          y=  "  E2. 

Then  x  —y  =  current  in  line. 

The  voltage  drop  across  the  points  AB  will  be  equal  along  the 
three  paths. 

Thus  voltage  across  AB  =2  —  .5x      =  (current  through  cell  EJ. 
1.4 +  .80  =(       "  "        E2) 

1.7 (x  —y)  —  (      "       resistance  R) 


FURTHER  APPLICATIONS  251 

Therefore 


^-X"'   .........     (1) 

and 


_2.2*-2 

Since  (1)  =  (2) 

.6 -.5*    2.2z-2 

.8 


.8 
=  .125  ampere. 

sc-y-l.00-.126 

=  .875  ampere. 

Thus  current  through  7^  =  1.00  ampere. 
#,  =  .125     " 
R  =  .875    " 

Voltage  across  AB  =  2—  .5#; 
«2T-;5; 
—  1.5  volts. 
Check 

Voltage  across  A R  =  (current  through  7?)  X  (resistance  of  /£); 
-.875X1.7; 
=  1 .49  volts. 

From  the  results  of  our  computation  we  see  that  the  current 
was  backing  up  through  cell  E2  on  account  of  its  low  E.M.F. 
This  shows  what  is  likely  to  happen  in  a  battery  of  storage  cells 
when  one  cell  or  set  of  cells,  joined  in  parallel  with  others,  becomes 
worn  out  before  the  rest. 

It  is  common  practice,  as  we  have  seen,  to  operate  shunt  and 
compound  generators  in  parallel.  The  necessity  for  using  machines 
of  approximately  equal  E.M.F.  and  resistance  is  apparent  from 
an  inspection  of  this  example.  The  distribution  of  current  in  line 


252  ELEMENTS  OF  ELECTRICITY 

and  generators  so  used,  is  similar  to  that  in  the  case  of  battery 
cells.  Thus,  if,  in  above  example,  Generators  be  substituted  for 
Battery  Cells,  the  method,  computation,  and  results  would  be 
the  same. 

Problem  12-9.  Assume  the  following  values  for  Fig.  193  and 
find  current  through  cells  and  line. 

EI  has  2.2  volts  E.M.F.  and  .4  ohm  internal  resistance. 
E2  has  1.8  "     .6 

R  has  .2  ohm  resistance. 

.  Problem  13-9.  A  series  set  of  5  cells  each  having  1.4  volts 
E.M.F.  and  .25  ohm  internal  resistance,  is  joined  in  parallel  to 
a  series  set  of  6  cells  of  1.3  volts  E.M.F.  and  an  internal  resistance 
of  .2  ohm  each.  The  parallel  combination  feeds  a  line  having 
a  resistance  of  .5  ohm. 
Find: 

(a)  Current  through  .5  ohm  resistance. 

(b)  Voltage  across  .5  ohm  resistance. 

Problem  14-9.  A  shunt  generatorof  120voltsE.M.F.and  .2ohm 
armature  resistance  is  joined  in  parallel  to  another  shunt  generator 
of  115  volts  E.M.F.  arid  .4  ohm  armature  resistance.  The  parallel 
combination  is  arranged  to  feed  a  line  having  10  ohms  resistance. 

Find: 

(a)  Current  through  line  and  through  each  generator. 

(6)  Voltage  across  line. 

155.  Efficiency  of  Direct  Current  Machines.  Stray 
Power.  The  efficiency  of  any  machine  or  any  combination 
of  machines  has  been  defined  as  the  ratio  of  the  OUTPUT 
to  the  INPUT.  As  an  equation  it  is  written  as  follows: 

T1~  .  output 

Efficiency  =—  —  —  . 
input 

In  generators  it  is  difficult  to  measure  the  input.  So  the 
output  and  losses  are  measured.  The  input  equals  the 
output  plus  the  losses.  The  equation  for  efficiency  of  a 
generator  then  becomes: 

Efficiency  = 


. 
input      output  +losses 


FURTHER  APPLICATIONS  253 

In  motors  it  is  easier  to  measure  the  input  and  the  losses. 
The  output  equals  the  input  minus  the  losses.  The  equation 
for  the  efficiency  of  a  motor  then  becomes: 

T,,-  .  output      input  — losses 

Efficiency  =  -=—   — ~  =— —. . 

input  input 

The  losses  in  a  machine,  whether  running  as  a  generator 
or  a  motor  may  be  divided  into  three  parts: 

(1)  Copper  loss  =  (I2R)  in  armature  and  field  wind- 
ings. 

(2)  Iron   losses  =  (hysteresis    and    eddy  currents    in 
armature  core) ; 

(3)  Mechanical  losses  =bearing  friction,   brush  fric- 
tion, and  windage. 

The  2d  and  3d  parts,  i.e.,  iron  losses  and  mechanical 
losses,  are  difficult  to  measure  separately.  For  this  reason 
they  are  generally  included  in  one  term  and  spoken  of  as 
the  STRAY  POWER  Loss. 

A  machine,  then,  may  be  said  to  have  but  two  losses. 

(1)  Copper  loss. 

(2)  Stray  power  loss. 

It  will  be  noted  that  all  the  losses,  of  which  the  STRAY 
POWER  is  composed,  depend  chiefly  upon  the  speed  of  the 
machine,  and  the  strength  of  the  magnetic  field. 

It  can  be  shown  also  that,  within  narrow  limits,  the  stray 
power  of  any  machine  is  nearly  proportional  to  the  voltage 
induced  in  the  armature  (which,  we  have  seen,  depends 
upon  the  speed  and  magnetic  field) .  Thus  if  the  field 
is  maintained  at  a  constant  strength  and  the  speed  of 
rotation  not  changed,  the  stray  power  of  either  a  motor 
or  a  generator  is  practically  constant.  Since  the  field 
of  a  shunt  machine  can  be  kept  constant  at  all  loads,  the 
stray  power  can  usually  be  measured  at  no  load  and  assumed 
to  remain  constant  when  the  machine  is  loaded,  provided 
the  speed  does  not  change. 


254  ELEMENTS  OF  ELECTRICITY 

If  either,  or  both  the  field  strength  and  speed  change 
with  the  load,  the  stray  power  at  any  load  can  be  found 
by  a  direct  proportion  between  the  stray  power  and  induced 
E.M.F.  at  the  different  loads.  This  is  approximately  true 
only  when  speed  change  is  slight,  say  under  10  or  12  per 
cent. 

156.  Method  of  Finding  Stray  Power  Loss  in  Shunt 
Dynamo.  Suppose  a  shunt  machine  is  run  unloaded,  as 
a  separately  excited  motor,  at  any  desired  speed.  Since 
the  machine  is  doing  no  work,  all  the  power  taken  by  the 
armature  is  lost.  This  power  then  takes  in  both  the 

(1)  Copper  loss  and 

(2)  Stray  power  loss. 

The  copper  loss  can  be  computed  by  getting  the  product 
of  the  armature  resistance  by  the  square  of  the  armature 
current.  If  now  this  computed  copper  loss  be  subtracted 
from  the  total  power  received  by  the  armature  the  remainder 
is  the  stray  power  loss.  This  may  be  stated  by  an  equation 
as  follows: 

P  —  EJ  —  7  27? 

*s  — &1  a       L  a  *-*>ai 

when  Ps  =  Stray  power  loss  in  watts. 

E  =  Impressed  voltage  across  the  armature. 
7a=Current  through  armature. 
R a  =  Resistance  of  armature. 

Example  (1).  A  shunt  dynamo  is  run  unloaded  as  a  motor 
with  the  field  separately  excited.  The  pressure  across  the  arma- 
ture is  120  volts,  the  current  through  armature  is  3.5  amperes. 
Armature  resistance  is  .2  ohm.  What  is  Stray  Power  Loss  at 
this  speed? 

The  total  power,  and,  in  this  case,  the  total  loss, 

=  EIa 
=  120X3.5 
=  420  watts. 
Copper  loss  =  IazRa 

=  3.52X.2 
=  2.45  watts. 


FURTHER  APPLICATIONS  255 

Stray  power  loss  =  Total  loss— copper  loss 
-420-2.5 
=  41 7.5  watts. 

Example  (2).  Assume  that  the  above  dynamo,  when  run  as 
a  shunt  generator  at  same  speed  and  field  strength,  delivers  50 
amperes  at  110  volts.  Field  resistance  =  55  ohms.  What  are  the 
total  losses? 

Solution.  Since  same  speed  and  field  are  maintained,  the 
induced  E.M.F.  in  armature  must  be  the  same.  Thus  the  stray 
power  is  constant. 

Total  losses  =  stray  power  +  copper  loss. 

Stray  power  =  41 7.5  watts. 

Copper  loss  in  field  =  Ij*Rf 

110 

//=-—-  =2  amperes. 
55 

//#/= 22X  55 

=  220  watts. 

Copper  loss  in  armature  =  Ia2Ra 
=  522X.2 
=  541  watts. 

Total  loss  =  541 +220  +  418 

=  1179  watts 
=  1.18K.W. 

Example  (3) .  If  the  generator  ran  at  sufficient  speed  to  supply 
50  amperes  at  114  volts,  what  would  total  losses  be? 
Solution : 

114 

Current  in  field  =  — —  =  2.07  amperes. 
55 

Current  in  armature =50 +  2.07  =  52.1  amperes. 
Stray  power  losses  in  this  and  previous  example  are  proportional 
to  the  values  of  the  induced  E.M.F.  since  the  variation   is  not 
great. 

Induced  E.M.F.  of  a  generator  equals  terminal  voltage  +  voltage 
required  to  force  current  through  the  armature. 
Induced  E.M.F.  in  example  (2)=E+IaRa; 

=  110  +  52X.2 
=  120.4 


256  ELEMENTS  OF  ELECTRICITY 

Induced  E.M.F.  in  example  (3)=E  +  IaRa 

=  114  +  52.1  X.2 
=  124.4. 

Ps  (at  114  volts)  _PS  (at  1  14  volts)  _  124.4 
Ps  (at  110  volts)"  417.5  ~12O4' 


=  430  watts. 

Copper  loss  in  field  =  ///?/ 

=  2.072X55 
=  237  watts. 

Copper  loss  in  armature  =  la~Ra 

=  52.12X.2 
=542. 

Total  loss  =  copper  loss  in  armature  +  copper  loss  in  field 

+  stray  power  loss 
=430  +  237  +  542 
=  1209  watts 
=  1.21K.W. 

When  it  is  possible,  the  usual  method  of  obtaining  the  stray 
power  loss  of  a  shunt  generator  is  to  run  the  machine  unloaded 
as  a  separately  excited  shunt  motor. 

The  voltage  impressed  across  the  armature  is  made  equal  to 
the  E.M.F.  which  the  machine  generates  when  supplying  full 
load  current,  i.e.,  the  terminal  voltage  as  generator  at  full  load 
plus  the  voltage  lost  in  armature  at  full  load.  The  field  strength 
is  then  adjusted  until  the  machine  runs  at  full  load  speed.  The 
total  input  into  the  armature  is  then  equal  to  the  stray  power 
at  full  load  plus  the  small  /02/?a  loss  of  no  load  current. 

Problem  15-9.  A  shunt  generator  is  run  as  a  separately  excited 
motor  at  no  load  with  speed  equal  to  full  load  generator  speed, 
and  impressed  voltage  equal  to  full  load  generator  E.M.F. 

Current  taken  by  armature  =  4.5  amperes. 

Impressed  voltage  across  armature  =  115  volts. 

Armature  resistance  =  .08  ohm. 

What  is  Stray  power  loss? 

Problem  16-9.  The  field  of  generator  in  Problem  15  has  a  resist- 
ance of  100  ohms.  What  is  total  loss  at  full  load  of  62.5  amperes 
at  110  volts? 


FURTHER   APPLICATIONS  257 

Problem  17-9.  If  generator  in  Problem  15  were  run  at  such  a 
speed  that  it  delivered  75  amperes  at  110  volts,  what  would  stray 
power  loss  be? 

156.  Generator   Efficiency;    Commercial   and  Electrical. 

The  efficiency  of  a  generator  may  be  stated  in  two  ways: 
either  as  the  COMMERCIAL  EFFICIENCY  or  as  the  ELEC- 
TRICAL EFFICIENCY.  The  COMMERCIAL  EFFICIENCY  is  the 
true  efficiency,  since  it  is  the  ratio  of  the  total  output  to 
the  total  input.  That  is 

.  .     £.".-,.  Output 

Commercial  efficiency  =  -rr—A — ,   ,  ^  ,   ,  . = 

Output  +  Total  losses 

Output     

Output  +  Copper  loss  +  Stray  power  loss* 

The  commercial  efficiency  of  generator  in  example  (2),  page 
255,  then  equals 

Output  (50X110) 

Output -f  Copper  loss  +  Stray  power  loss  ~  (50  X 1 10)  + 1 180 

_5500 
"6680 
=  82.4  per  cent. 

The  ELECTRICAL  EFFICIENCY  is  the  ratio  of  the  electrical 
output  to  the  electrical  power  generated.  The  electrical 
power  generated  equals  the  electrical  power  delivered 
to  line,  plus  electrical  power  lost .  in  armature  and  field 
coils;  that  is,  plus  the  copper  loss.  Thus  the 

Electrical   efficiency  =^— - —     — ~ —  — . 

Output  +  Copper  losses 

The  electrical  efficiency  of  generator  in  example  (2),  page 
255,  then  equals: 

Output 50X110 

Output  + Copper  losses  ~  (50  X 1 10)  +  (220  +  541) 
^5500 
"6261 
=  88  per  cent, 


258  ELEMENTS  OF  ELECTRICITY 

Unless  otherwise  stated,  the  COMMERCIAL  EFFICIENCY 
of  a  generator  is  understood  by  a  statement  of  its  efficiency. 

Problem  18-9.  What  is  the  commercial  efficiency  and  the 
electrical  efficiency  of  generator  in  Problem  16? 

Problem  19-9.  What  is  the  commercial  and  the  electrical 
efficiency  of  generator  in  Problem  17? 

Problem  20-9.  What  is  the  electrical  efficiency  of  generator 
in  Problem  7? 

Problem  21-9.     A  50  K.W.  generator  has  a  stray  power  loss 
at  full  load  of  1500  watts,  and  a  copper  loss  of  1800  watts.     What 
\  is  the  electrical  and  the  commercial  efficiency? 

158.  Motor    Efficiency.     Commercial    and    Mechanical. 

The  efficiency  of  a  motor  may  be  stated  in  two  ways: 
either  as  the  COMMERCIAL  EFFICIENCY,  or  as  the  MECHAN- 
ICAL EFFICIENCY.  The  commercial  efficiency  is  the  true 
efficiency,  since  it  is  the  ratio  of  the  total  output  to  the 
total  input.  That  is,  of  a  motor,  the 

.  ,      ~  .  Input  —  Total  losses 

Commercial   efficiency  =  —      —  ^  --  -  —      —  . 

Example.  The  stray  power  of  a  shunt  motor  at  full  load  speed 
and  voltage  =  400  watts.  The  armature  resistance  =  .02  ohm. 
The  field  resistance  =  55  ohms.  When  running  at  full  load  the 
motor  takes  62  amperes  at  110  volts.  What  is  the  commercial 
efficiency? 

Input—  Total  losses 

Commercial  efficiency  =  -  —  . 

Input 

Input  =  62  X  1  10  =  6820  watts. 
Copper  losses  : 

In  Armature  =602X.02 
=  72  watts. 

In 


-220  watts. 

Total  copper  loss  =  292  watts. 
Stray  power      "    =400     " 
Total  losses  =692    " 


FURTHER  APPLICATIONS  259 

6820-692 
Commercial   efficiency  = 

OdrfwVJ 

^6128 
~6820 

=89.8%. 

The  MECHANICAL  EFFICIENCY  is  ratio  of  the  Mechanical 
Output  to  the  Mechanical  Power  Developed  in  armature. 
The  Mechanical  Power  Developed  equals  all  the  power  not 
used  in  overcoming  electrical  resistance  of  field  and  arma- 
ture. 

That  is,  total  power  developed  in  armature  equals 
Input  —  Copper  losses,  or 
Output  +  Stray  power  losses. 

Input  — Total  losses 
Mechanical  efficiency  =- — 

Input— Copper  losses 

or 

Output 


Output  +  Stray  power 

Example.     What    is   the    Mechanical  Efficiency   of   motor   in 
above  example? 

Input  —Total  losses 
Mechanical  efficiency  =  -  --  -  —  ^  —    —  ;  — 

Input—  Copper  losses 

_  (62X110)  -692 
"(62X110)  -292 


or 


6528 
=  93.8%; 
Output 


Output  +  Stray  power 

(62X110) -692 
"  (62X110)  -692  +  400 
_6128 
~6528~ 
^93,8%. 


260  ELEMENTS  OF  ELECTRICITY 

Unless  otherwise  stated,  the  COMMERCIAL  EFFICIENCY  is 
meant  by  the  term  efficiency  when  applied  to  a  motor. 

Problem  22-9.  (a)  What  is  the  commercial  efficiency  of  motor 
in  Problem  34-8  if  the  stray  power  loss  is  600  watts?  (6)  What 
is  the  mechanical  efficiency? 

Problem  23-9.  Assume  the  stray  power  loss  in  motor  of  Prob- 
lem 40-8  to  be  700  watts.  What  is  the  mechanical  and  the  com- 
mercial efficiency? 

Problem  24-9.  Mechanical  efficiency  of  motor  in  Problem 
41-8  is  92  per  cent.  What  is  the  stray  power  loss? 

Problem  25-9.  The  commercial  efficiency  of  motor  in  Problem 
47-8  is  90  per  cent,  (a)  What  is  stray  power  loss?  (b)  What  is 
mechanical  efficiency? 


FURTHER  APPLICATIONS  261 


SUMMARY  OF  CHAPTER  IX 

ELECTRICAL  TRANSMISSION.  Is  convenient  and  effi- 
cient over  long  distances. 

LINE  LOSS.  Varies  inversely  as  the  square  of  the  voltage 
of  transmission.  For  a  given  line  loss,  the  size  of  the  wire, 
either  by  cross  section  or  by  weight,  varies  inversely  as  the 
voltage. 

FEEDERS  are  used  in  low  voltage  systems  to  increase 
size  of  conductors.  Power,  voltage  and  current  distribu- 
tion solved  by  Ohm's  Law. 

THREE-WIRE  SYSTEM.  Neutral  wire  intended  to  carry 
very  little  current. 

Requires  about  three -eighths  as  much  wire  as  a  two-wire 
system  of  same  capacity,  and  of  same  voltage  as  one  leg. 

Affords  two  standard  voltages. 

KIRCHHOFF'S  LAWS.  (An  extension  of  Ohm's  Law.) 
(ist)  Sum  of  currents  flowing  away  from  a  point  in  a  circuit 
equals  sum  of  currents  flowing  to  the  point.  (2d)  In  any 
circuit,  algebraic  sum  of  E.M.F.'s  equals  sum  of  IR  drops. 
In  a .  circuit  containing  no  source  of  E.M.F.,  sum  of  IR 
drops  clockwise  around  the  circuit,  equals  sum  of  IR  drops 
counter  clockwise. 

Are  applied  to  find  current  distribution  in  "  mesh  work  " 
and  in  systems  of  generators  or  battery  cells  operated  in 
parallel. 

POWER  LOSS  IN  DYNAMOS.  There  are  three  sources  of 
loss  in  electrical  machines,  whether  run  as  generators  or 
motors : 

(1)  Copper  loss,  (PR  in  Armature  and  Field.) 

(2)  Iron  loss   (Hyteresis  and  Eddy  Currents.) 

(3)  Mechanical   losses    (Windage,   Friction,   etc.) 
Items  (2)  and  (3),  i.e.,  iron  loss  and  mechanical  losses,  are 

generally  grouped  as  the  stray  power  loss. 
EFFICIENCY     OF     GENERATORS. 

^  Output  Output 

Commercial  Efficiency  = 


Electrical  Efficiency        = 


Input       Output  +  Total  Losses 

Output 

Electrical  Power  Developed 

Output 

Output  -I-  Copper  Losses' 


262  ELEMENTS   OF   ELECTRICITY 

EFFICIENCY  OF  MOTORS. 

Output      Input  —Total  Losses 

Commercial  Efficiency  =  -^r— —  — . 

Input  Input 

Output 
Mechanical  Efficiency 


Mechanical  Power  Developed 

Input  -Total  Losses Output 

Input  —Copper  Losses     Output  -f  Stray  Power" 


PROBLEMS   ON  CHAPTER  IX 

26-9.  A  500  K.W.  generator  is  to  be  selected  to  transmit  power 
6  miles  over  a  No.  0  copper  wire.  Plot  a  curve  showing  relation 
of  voltage  to  efficiency  of  transmission  for  above  power,  ranging 
from  0  per  cent  efficiency  to  99  per  cent.  Have  at  least  12 
points  about  equally  spaced  along  the  curve. 

27-9.  State  what  voltage  you  would  select  to  transmit  power 
in  Problem  26.  Using  3  times  this  voltage,  what  size  wire  could  be 
used  with  the  same  line  loss? 

28-9.  Allowing  5  per  cent  of  the  power  generated  to  be  lost  in 
the  line,  plot  a  curve  between  size  of  the  copper  wire  (in  circular 
mils)  and  voltage  of  transmission  in  Problem  26.  Plot  at  least  8 
points. 

29-9.  What  size  aluminum  wire  could  be  used  in  Problem  27, 
with  same  line  loss? 

30-9.  What  size  copper  wire  is  required  to  deliver  current 
at  110  volts  to  a  25  H.P.  motor  of  90  per  cent  efficiency,  if  the 
motor  is  1500  feet  from  the  115-volt  generator? 

31-9.  Generator  armature  of  Problem  30  has  a  resistance  of  .02 
ohm.  (a)  What  is  its  E.M.F.?  (6)  What  is  its  electrical  efficiency? 
(c)  What  is  the  efficiency  of  transmission? 

32-9.  An  electric  railway  is  in  the  form  of  a  rectangle  6  miles 
by  2 2  miles.  The  generator  station  is  in  the  middle  of  one  of 
the  long  sides.  The  trolley  wire  is  No.  0  hard-drawn  copper. 
The  track  resistance  is  .04  ohm  per  mile.  There  is  one  No.  0000 
feeder  running  from  generator  station  directly  across  rectangle 
to  the  other  side,  where  it  is  joined  to  the  trolley  wire.  There 
are  2  cars  on  line  at  a  given  instant,  one  in  middle  of  each  short 


FURTHER   APPLICATIONS  263 

side  of  rectangle;  one  is  taking  100  amperes,  the  other  60  amperes. 
What  is  the  voltage  across  each  car,  if  the  brush  potential  of  the 
generator  is  570  volts? 

33-9.  If  there  were  4  cars  on  the  line  in  Problem  32,  one  at 
each  corner  of  the  rectangle,  each  taking  60  amperes,  what 
would  voltage  be  across  each  car? 

34-9.  Suppose  feeder  in  Problem  32  broke,  what  would  be 
the  drop  across  each  car? 

35-9.  A  9-mile  trolley  line  with  generator  station  at  one 
end  has  two  feeders.  One  of  No.  0000  soft  copper  wire  extends 
from  generator  6  miles  along  trolley  and  is  tied  to  trolley  every 
2  miles.  The  other  of  hard-drawn  copper,  800,000  C.M.  in  cross- 
section,  extends  from  generator  along  the  trolley  for  3  miles  and 
is  tied  to  trolley  at  the  end  only.  There  are  3  cars  on  the  line, 
distributed  as  follows:  Car  I,  3  miles  from  generator  station, 
takes  100  amperes.  Car  II,  5  miles  from  station,  takes  50  amperes. 
Car  III,  7  miles  from  station,  takes  40  amperes.  Trolley  wire 
is  No.  0  hard-drawn  copper.  Track  resistance  is  .03  ohm  per 
mile.  Generator  voltage  is  580  volts,  (a)  What  is  voltage 
across  each  car?  (6)  What  is  efficiency  of  transmission? 

36-9.  If  trolley  wire  in  Problem  35  breaks  between  generator 
and  place  where  No.  0000  feeder  is  tied  to  it,  what  would  voltage 
across  each  car  become? 

37-9.  The  shunt  motor  of  a  motor-generator  set  takes  60  amperes 
at  110  volts.  The  stray  power  loss  of  motor  is  430  watts.  Field 
resistance  is  100  ohms.  Armature  resistance  is  .08  ohm.  The 
stray  power  loss  in  generator  is  420  watts;  armature  resistance 
=  .0008  ohm.  Shunt  field  resistance,  10  ohms.  Electrical 
efficiency  of  generator  93  per  cent.  What  is  terminal  voltage 
of  generator,  if  it  is  supplying  600  amperes? 

38-9.  (a)  What  is  the  commercial  and  the  mechanical  efficiency 
of  motor  in  Problem  37?  (b)  What  is  the  commercial  efficiency  of 
the  generator?  (c)  What  is  the  efficiency  of  the  set? 

39-9.  A  shunt  generator,  the  armature  of  which  has  a  resistance 
of  .5  ohm  is  charging  a  set  of  40  storage  cells  in  series.  E.M.F. 
of  generator  =  120  volts.  Each  cell  has  a  resistance  of  .002  ohm 
and  an  E.M.F.  of  2.5  volts.  Allow  .15  ohm  for  lead  wire,  connec- 
tions, etc.  Neglect  current  through  field  coils,  (a)  What  current 
is  generator  delivering?  (b}  What  is  terminal  voltage  of  genera- 
tor? (c)  What  is  voltage  across  each  cell? 


264  ELEMENTS  OF  ELECTRICITY 

40-9.  Each  of  two  shunt  generators  has  an  E.M.F.  of  120  volts 
and  an  armature  resistance  of  .08  ohm.  If  the  two  generators 
are  joined  in  series  across  a  line  of  10  ohms  resistance,  what  current 
will  flow?  Neglect  field  currents. 

41-9.  If  generators  of  Problem  40  are  joined  in  parallel  across 
same  line,  what  current  will  flow?  Neglect  field  currents. 

42-9.  Assume  the  field  resistance  of  each  generator  of  Problem 
40  to  be  60  ohms,  (a)  What  is  the  brush  potential  of  each  when 
joined  as  in  Problem  40?  (6)  When  joined  as  in  Problem  41? 

43-9.  Assume  one  of  the  generators  in  Problem  40  had  an  E.M.F- 
of  115  volts  and  an  armature  resistance  of  .06  ohm,  and  other 
remained  as  in  Problem  40.  What  current  would  flow  through  line 
if  joined  as  in  Problem  40?  Neglect  field  currents. 

44-9.  If  generators  of  Problem  43  were  joined  as  in  Problem  41, 
what  current  would  flow?  Neglect  field  currents. 

45-9.  An  electric  arc  between  carbons  has  a  counter  E.M.F. 
of  33  volts.  When  42  volts  are  impressed  across  the  arc,  and 
9.5  amperes  flow,  what  resistance  has  arc? 

46-9.  It  is  desired  to  light  a  house  with  300  incandescent  lamps, 
in  parallel.  The  lamps  have  a  hot  resistance  of  200  ohms  and 
operate  on  110  volts.  A  battery  of  storage  cells  is  to  be  used, 
each  of  which  has  an  E.M.F.  of  2.2  volts,  an  internal  resistance 
of  .004  ohm,  and  normal  discharge  rate  is  30  amperes.  How  many 
cells  are  required  and  how  should  they  be  arranged? 

47-9.  A  motor  being  tested  by  a  Prony  brake,  runs  at  1200 
R.P.M.,  and  takes  67  amperes  at  112  volts.  Brake  arm  is  16  inches 
long,  balance  registers  28  Ibs.  What  is  input,  output,  arid  efficiency 
of  motor? 

48-9.  Assume  that  each  lamp  in  Fig.  194  takes  2  amperes,  and 
that  resistance  of  the  lamps  remains  constant.  Brush  potential  of 
each  generator  is  110  volts.  Find: 

(a)  Line  drop  in  each  section. 

(6)  Voltage  across  each  set  of  lamps. 

(c)  Power  delivered  by  Gl  and  G2. 

(d)  Efficiency  of  transmission. 

49-9.  If  a  break  occurs  in  neutral  between  0  and  S,  what  would 
be  the  values  of  (a),  (6),  (c)  and  (d),  Problem  48-9? 

50-9.  If  a  break  occurs  in  the  neutral  between  S  and  V  what 
would  be  the  values  of  (a),  (6),  (c)  and  (d),  Problem  48-9? 


FURTHER  APPLICATIONS 


265 


51-9.  A  building  is  supplied  by  a  two-wire  system,  the  wiring 
diagram  of  which  is  as  per  Fig.  195.  Values  on  lines  represent 
the  resistance  of  that  section  of  line  wire.  Motor  M  takes  40 
amperes.  Each  lamp  takes  4  amperes,  and  resistance  of  each 


.2  Ohm 


.20hi 


/N 


.2  Ohm               s 

1 

.2  Ohm 

'             °ffi 

33 

.4  Ohm 
FIG.  194. 


is  constant,  (a)  Find  voltage  across  each  set  of  lamps.  (6) 
Draw  diagram  indicating  amount  and  direction  of  current  in  each 
section  of  line. 

NOTE.  The  IR  drop  across  NMR  must  equal  IR  drop  across 
NSR,  since  these  are  parallel  circuits  between  the  same  two  points 
R  and  N.  Similarly  treat  points  P  and  T. 


.04 


.04 


.04  .04 

FIG.  195. 


.05 


To  Gen. 
125  Volte 


.05 


52-9.  Find   the   voltage   across   the   various  sets  of  lamps  of 
Problem  51-9,  when  the  motor  is  not  running. 

53-9.  Repeat  Problem  52-9,  leaving  out  jumpers  MN  and  OP. 

54-9.  For  running  a  coil  of  2  ohrns  resistance,  4  storage  cells 
in  parallel  are  used.     Each  had,  when  new,  an  E.M.F.  of  2.4  volts 


266  ELEMENTS  OF  ELECTRICITY 

and  internal  resistance  of  .04  ohm.     What  current  flows  through 
the  line  in  this  case,  and  what  is  terminal  voltage  of  the  battery? 

55-9.  One  of  the  cells  of  Problem  54  was  damaged.  Its  E.M.F. 
fell  to  2.0  volts  and  its  internal  resistance  rose  to  .08  ohm.  The 
four  cells  are  still  used  as  in  Problem  54.  (a)  What  is  current  in 
line?  (6)  Terminal  voltage  of  battery? 

56-9.  A  shunt  motor  when  delivering  full  load  takes  40  amperes 
at  112  volts.  When  running  with  no  load  at  same  speed  it  takes 
3  amperes  at  106  volts.  Field  resistance  is  100  ohms  and  armature 
resistance  is  .125  ohm.  (a)  What  power  does  it  deliver  at  full 
load?  (6)  What  are  the  commercial  and  the  mechanical  efficiencies? 

57-9.  In  system  arranged  as  in  Fig.  192,  each  lamp  takes  10 
amperes.  The  resistance  of  the  different  sections  of  the  line  is  as 
follows  : 

A  B  =  .05  ohm.  CD  =  .06  ohm. 

BC  =  .  04     "  EF=.Q3     " 

M     "  ^  =  .02     " 


E.M.F.  of  (^  =  130  volts.  E.M.F.  of  <72  =  135  volts. 

Find: 

(a)  Voltage  across  group  I  and  across  group  II. 
(6)  Terminal  voltage  of  Gl  and  (?2. 

58-9.  Find  efficiency  of  transmission  in  Problem  57-9. 

59-9.  The  generators  in  Problem  57  are  of  the  shunt  type.  The 
field  of  G^  has  58  ohms  resistance,  of  G?,  62  ohms.  What  is  the 
electrical  efficiency  of  each? 

60-9.  What  would  the  total  losses  be  in  generator  of  Problem 
15-9  at  full  load  with  terminal  voltage  of  112  volts?  Field=100 
ohms. 

61-9.  A  5-K.W.  110-  volt  compound  generator,  of  long  shunt 
type,  has  stray  power  loss  at  full  load  of  600  watts.  Resistance 
of  series  field  =  .025  ohm,  of  armature  =  .015  ohm,  of  shunt  field, 
55  ohms.  Find  the  electrical  and  the  commercial  efficiencies. 

62-9.  If  same  data  as  in  Problem  61-9  held  for  a  short  shunt 
generator,  what  would  the  two  efficiencies  be? 

63-9.  If  machine  in  Problem  61  is  used  as  a  motor  and  takes 
48  amperes  at  112  volts  at  full  load,  what  will  the  commercial  and 
the  mechanical  efficiencies  be? 

64-9.  If  data  of  Problem  63  held  for  a  short  shunt  motor,  what 
would  the  two  efficiencies  be? 


FURTHER  APPLICATIONS  267 

65-9.  In  Fig.  106  the  terminal  voltage  of  the  battery  cell 
is  2  volts.  Resistance  Rl=5  ohms,  R2=S  ohms,  7£3  =  10  ohms, 
R  -=12  ohms.  Galvanometer  line  is  assumed  closed,  the  resist- 
ance of  which  is  25  ohms.  What  current  is  flowing  in  each  section 
of  unbalanced  bridge? 

66-9.  According  to  Kelvin's  Rule  for  most  economical  size 
conductor,  what  size  copper  wire,  B.  &  S.  gauge,  should  be  used 
under  the  following  conditions? 

Cost  of  installed  copper  wire  per  lb.,  $.20. 

Cost  of  generating  power  $.01  per  K.W.  hr. 

Probable  rate  of  interest,  5  per  cent. 

Power  to  be  transmitted,  1000  K.W. 

Voltage  of  transmission,  550  volts. 

Power  to  be  used  10  hrs.  per  day  throughout  year. 

67-9.  Compute  size  of  aluminum  wire  that  should  be  used 
under  conditions  of  Problem  66,  assuming  that  aluminum  costs 
twice  as  much  per  lb.  installed  as  copper. 

68-9.  If  double  the  voltage  of  Problem  66-9  is  used  what  size 
copper  conductor  is  the  most  economical? 


CHAPTER  x 

INDUCTANCE 

Mutual  Inductance — Cause  of;  Effect  of;  Lenz's  Law — Induction 
Coils;  Jump  Spark;  Ruhmkorff — Transformers — Self  Induct- 
ance; Cause  and  Effect  of — Induction  Coil;  Make  and  Break — 
Inductance,  a  Property  of  the  Circuit;  Unit  of  Inductance,  the 
Henry — Computation  of  Self  Inductance;  of  Mutual  Inductance; 
of  Inductance  in  Transmission  Lines — Effect  of  Inductance  in 
Alternating  Current  Circuits. 

THE  exact  nature  of  magnetism  and  electricity  is  still 
unknown,  but  much  has  been  discovered  concerning  the 
relations  existing  between  them. 

We  have  seen  that  wherever  there  is  a  current  of  electricity 
flowing,  there  is  always  a  magnetic  field  present.  The 
strength  of  this  magnetic  field  is  proportional  to  the  strength 
of  the  current,  and  the  direction  of  it  always  holds  the 
same  relation  to  direction  of  the  current. 

We  have  also  seen  that  when  an  electric  conductor  cuts 
the  lines  of  force  in  a  magnetic  field,  an  electric  pressure 
is  induced  in  the  wire  tending  to  cause  a  current  to  flow. 
The  strength  of  this  induced  voltage  is  proportional  to  the 
rate  of  cutting.  The  direction  of  it  has  a  definite  relation 
to  the  direction  of  the  field  and  to  the  direction  of  the  motion. 

We  have  further  seen  that  when  a  steady  current  is 
flowing  through  an  electric  circuit,  a  certain  definite  pressure 
is  required  to  keep  this  current  flowing.  This  pressure 
depends  upon  the  resistance  of  the  circuit  and  upon  the 
work  to  be  done  by  the  current.  That  is,  the  resistance 
of  the  conductors  must  be  overcome,  and  the  counter 

268 


INDUCTANCE 


269 


E.M.F.  of  any  revolving  motor  armature,  etc.;  just  as  in 
a  machine,  the  friction  of  the  bearings  must  be  overcome 
and  the  opposing  force  offered  by  any  work  that  is  being 
done.  As  long  as  the  current  in  an  electric  circuit  is  steady, 
we  have  seen  by  Ohm's  Law,  that  a  definite  steady  elec- 
tromotive force  is  required  to  maintain  it,  just  as  a  definite 
steady  mechanical  force  is  required  to  keep  a  machine 
turning  against  a  steady  resistance.  So  far,  we  have  been 
dealing  with  steady  currents, 
only.  Nothing  has  been  said 
about  the  force  required  in 
starting  and  stopping  a  flow  of 
electricity. 

195.  Inductance.  Induced 
E.M.F.  Suppose  we  have  two 
coils  of  wire,  A  and  B,  Fig.  196. 
Coil  A  is  placed  within  coil  B, 
but  has  no  electrical  connection 
to  it.  Across  the  terminals  of 
B  a  voltmeter  is  placed.  Now 
when  switch  S  is  thrown  to 
power,  a  current  rushes  into  coil 
A  and  flows  around  the  coil 
counter-clockwise,  as  marked. 
But  also,  strangely  enough,  a 
momentary  current  flows  around 
in  coil  B,  through  the  voltmeter, 

only  in  the  opposite  direction.     This  momentary  current  in 
B  lasts  but  for  an  instant,  and  dies  out. 

If  now,  we  suddenly  open  the  switch  S,  in  order  to  stop 
the  current  in  A,  we  notice  that  another  momentary  current 
is  set  up  in  B.  This  time  the  momentary  current  in  B  is 
in  the  same  direction  as  the  current  we  are  stopping  in  A. 
This  momentary  current,  like  the  first  one,  dies  out  almost 
instantly.  Thus  when  we  closed  the  switch  and  started 
a  current  in  A,  we  noticed  a  momentary  current  set  up  in 


FIG.  196. — Induction  coils. 
Mutual  induction. 


270  ELEMENTS  OF  ELECTRICITY 

B,  opposite  in  direction  to  the  current  we  were  starting 
in  A.  On  the  other  hand,  while  we  were  stopping  the  cur- 
rent in  A,  we  noticed  a  momentary  current  set  up  in  B, 
in  the  same  direction  as  the  one  we  were  stopping.  These 
momentary  currents  set  up  in  B,  whenever  there  was  any 
change  in  the  current  in  A,  are  called  INDUCED  CURRENTS. 
The  E.M.F.  tending  to  cause  them  to  flow  is  called  an 
INDUCED  E.M.F.  and  the  two  electric  circuits  are  said  to 
possess  the  property  of  MUTUAL  INDUCTANCE. 

Note  that,  though  in  one  case  the  induced  current 
was  in  the  opposite  direction  to  the  current  in  A,  and  in 
the  other  case  in  the  same  direction,  in  both  cases  it  opposed 
the  change  taking  place  in  the  current  in  A.  It  opposed 
the  setting  up  of  a  current  in  A,  when  we  threw  the  switch, 
by  setting  up  one  in  the  opposite  direction.  When  we 
pulled  the  switch,  it  opposed  the  stopping  of  the  current 
in  A,  by  setting  up  one  in  the  same  direction  as  the  one  we 
were  stopping.  In  each  case  as  soon  as  the  change  in  the 
current  in  A  was  over,  the  induced  current  died  out. 

Let  us  consider  also  the  magnetic  field  set  up  by  these 
induced  currents  in  B.  When  we  sent  the  current  into 
A,  we  were  setting  up  a  field  within  the  coil  making  this 
end  a  north  pole.  Note  that  the  induced  current  in  B 
opposed  this  setting  up  of  a  north  pole  by  setting  up  a  field 
with  a  south  pole  at  this  end,  in  order  to  neutralize  it. 
But  after  the  field  was  once  set  up,  the  current  in  B  ceased 
trying  to  neutralize  it  and  stopped  flowing.  As  soon, 
however,  as  we  began  to  destroy  this  field  by  stopping  the 
current  in  A,  a  current  was  induced  in  B,  which  opposed 
the  dying  out  of  the  field,  by  setting  up  a  field  of  its  own  in 
the  same  direction  as  the  one  we  were  destroying.  In  each 
case  the  field  of  the  induced  current  opposed  the  change 
that  was  taking  place  in  the  field  within  the  coils. 

All  these  phenomena  are  merely  manifestations  of  a  great 
law  first  stated  'by  Lenz,  and  called  Lenz's  Law,  part  of 
which  is: 


INDUCTANCE  271 

That  an  induced  current  is  always  in  such  a  direction 
that  its  field  opposes  any  change  in  the  existing  field. 

It  is  part  of  the  law  of  "  conservation  of  energy,"  as 
applied  to  the  electric  circuit. 

It  remains  to  be  shown  how  it  is  really  the  magnetic 
field  set  up  by  an  electric  current  which  causes  these  phe- 
nomena of  induction;  how  the  magnetic  field  contains 
the  kinetic  energy  of  a  circuit,  just  as  a  fly  wheel  and  other 
moving  parts  contain  the  kinetic  energy  of  a  machine  in 
motion. 

In  getting  a  machine  up  to  speed,  we  must  overcome 
not  only  the  force  of  friction  but  also  the  INERTIA  reaction 
during  the  process.  In  setting  up  a  current  in  a  circuit, 
we  must  overcome  not  only  the  resistance  (friction)  of 
the  circuit,  but  also  the  INDUCTANCE  (inertia)  reaction  of 
the  magnetic  field.  In  trying  to  stop  a  machine  which  has 
a  large  fly  wheel,  we  feel  conscious  of  some  reaction  tending 
to  keep  the  machine  turning,  i.e.,  an  opposing  force,  which 
increases  enormously  if  an  effort  is  made  to  stop  the  wheel 
suddenly.  This  is  the  reaction  due  to  the  inertia  of  the 
machine.  Similarly,  we  feel  conscious  that  there  is  a 
reaction  which  tends  to  oppose  the  sudden  stopping  of  an 
electric  current  which  has  a  strong  magnetic  field,  i.e.,  there 
is  a  force  which  tends  to  keep  such  a  current  flowing.  This 
is  the  reaction  due  to  the  INDUCTANCE  of  the  circuit.  The 
larger  the  MOMENT  OF  INERTIA  of  the  fly  wheel,  the  larger 
this  force  opposing  any  effort  to  stop  it  in  a  given  time. 
The  larger  the  INDUCTANCE  of  the  electric  circuit,  the  larger 
the  force  which  opposes  the  stopping  of  the  current  flowing 
through  it  in  a  given  time. 

Just  as  the  reaction  due  to  inertia  opposes  the  increasing 
or  decreasing  the  speed  of  a  fly  wheel,  so  the  reaction  due 
to  inductance  opposes  the  increasing  or  decreasing  of  the 
current  in  an  electric  circuit.  Let  us  investigate  the  causes 
and  the  practical  results  of  this  property  of  a  magnetic 
field. 


272 


ELEMENTS  OF  ELECTRICITY 


160.  Induction  Coils.  Mutual  Inductance.  In  Fig.  196, 
the  inner  coil  A  is  composed  of  few  turns  of  coarse  wire. 
If  a  current  were  sent  into  coil  A,  let  us  consider  what 
would  happen  while  the  current  was  building  up  to  its 
maximum  value.  Fig.  197  (6),  represents  a  vertical  cross 
section  along  the  axis  of  the  coil. 


(a) 


(6) 
FIG.  197. — Section  of  induction  coil. 


Fig.  197  (a),  represents  an  ideal  end  view.  Assume 
the  current  to  enter  coil  A  at  right-hand  end  and  to  flow 
in  Sit  the  top  of  the  loop  A  and  out  at  the  bottom  of  the 
loop  A.  This  would  cause  a  clockwise  field  to  grow  around 

the  top  wires  of  coil  A  and 
a  counter-clockwise  field 
around  the  bottom  wires. 
This  field  spreads  out  in 
ever  widening  rings  as  the 
current  is  increased  and 
cuts  across  the  sides  of  the 
coil  B.  This  is  shown  in 
Fig.  198(1),  (2),  (3),  where 
the  wire  A  represents  the  end  of  a  wire  in  coil  A,  and  B 
represents  the  end  of  a  wire  in  coil  B.  It  shows  that  as 
the  current  in  A  grows,  the  magnetic  field  caused  by  it, 
spreads  out  and  cuts  across  the  wires  of  coil  B.  The  wires 


FIG.  198. — Effect  of  growing  field  in  wires 
of  induction  coil. 


INDUCTANCE  273 

on  the  top  of  coil  B  are  cut  by  the  lines  as  they  move 
upward.  This  is  equivalent  to  the  wires  of  B  moving 
downward  and  cutting  the  lines  as  shown  in  Fig.  199.  By 
applying  the  right-hand  rule,  we  see  that  there  would  be  a 
voltage  induced  in  B  tending  to  send  the  current  OUT  of  B. 
This  is  in  the  opposite  direction  to  the  inducing  current  in  A. 

If  we  apply  this  to  the  lower  sides  of  the  coil,  Fig.  197, 
we  find  that  the  inducing  current  in  coil  A  is  out  in  the 
wires  (Ai)  at  the  bottom,  and  that  the  wires  BI  of  coil 
B  are  cut  by  the  growing  field  in  such  a  way  as  to  induce 
a  voltage  which  tends  to  send  a  current  in  at  BI.  Here 
again  we  see  that  the  induced  voltage  in  coil  B  is  opposed 


FIG.  199. — Direction  of  E.M.F.         FIG.  200. — End  view  of  induction  coil, 
induced  in  secondary  coil.  primary  current  growing. 

in  direction  to  the  increasing  current  in  coil  A.  If  we  look 
at  the  end  of  the  coil,  Fig.  300,  we  see  that  the  inducing 
electric  current  in  A  is  in  a  counter-clockwise  direction. 
As  long  as  it  is  increasing,  it  is  causing  an  induced  current 
to  flow  in  B  in  a  clockwise  direction. 

The  magnetic  field  in  the  air  core,  due  to  the  current  in 
B,  is  opposed  to  the  building  up  of  the  field  due  to  the  current 
in  A. 

The  current  in  A  is  trying  to  build  up  a  field  OUT  (Fig. 
200),  while  the  current  induced  in  B  tends  to  oppose  this 
building  up  of  a  field  OUT  by  neutralizing  it  with  a  field  IN. 

There  is  no  electrical  connection  between  coil  A  and  B. 
They  are  very  carefully  insulated  from  each  other.  Yet 


274  ELEMENTS  OF  ELECTRICITY 

all  the  time  that  a  current  is  increasing  in  one  direction 
in  A,  it  causes  a  current  to  build  up  in  B.  The  field  of  this 
induced  current  in  B  opposes  the  building  up  of  the  field 
by  the  current  in  A.  As  soon  as  the  current  in  A  reaches 
its  normal  value,  the  lines  of  force  around  the  wires  no  longer 
will  be  spreading  out  and  there  will  be  no  lines  cutting  the 
sides  of  B,  and  thus  the  induced  current  in  B  will  die  out. 

The  induced  current  thus  lasts  only  as  long  as  the  "  pri- 
mary "  current  (the  current  in  A)  is  growing.  While  the 
current  in  A  was  growing,  it  had  to  overcome  both  the 
resistance  of  the  wires  composing  the  coil,  and  also  the 
opposition  of  the  induced  current  in  5,  to  the  building  up 
of  the  field.  As  soon  as  the  current  in  A  reached  its  normal 
value,  the  induced  current  in  B  died  out,  and  there  was 
left  the  resistance  only  of  the  wires  of  A  to  be  overcome.  The 
opposition  to  the  building  up  of  the  field  is  due  to  the 
INDUCTANCE  of  the  coil.  Since  this  opposition  exists  in 
a  coil  not  connected  electrically  with  first  coil,  the  two 
coils  are  said  to. possess  MUTUAL  INDUCTANCE. 

As  long  as  a  force  is  trying  to  speed  up  a  machine,  it  has 
to  overcome  both  the  friction  (resistance)  and  a  certain 
opposition  to  the  building  up  of  speed.  As  soon  as  the 
normal  speed  is  reached,  the  opposition  to  building  up 
ceases,  and  there  is  only  the  friction  left  to  be  overcome. 
This  opposition  to  building  .up  the  speed  is  due  to  the 
MOMENT  OF  INERTIA  of  the  moving  parts,  as  we  have  stated 
before. 

By  comparing  these  two  statements,  it  can  be  seen  that 
there  is  a  close  analogy  between  inertia  and  inductance ,  in 
that  they  both  oppose  an  increase. 

But  inertia  reaction  not  only  opposes  any  increase  in 
speed,  but  also  opposes  any  decrease  in  speed.  Let  us 
now  consider  how  inductance  reaction  opposes  the  decrease 
of  the  current.  Fig.  201  (1),  (2),  (3),  represents  the  cur- 
rent in  A  dying  out.  The  field  around  each  wire  now 
contracts  and  in  so  doing,  again  cuts  the  wires  of  coil 


INDUCTANCE 


275 


B.  But  this  time  the  cutting  is  in  the  opposite  direction 
to  what  it  was  when  the  field  was  spreading  out.  As 
the  field  was  spreading  out,  there  was  a  current  of  OUT 
induced  in  the  coil  J3,  which  was  opposite  to  the  current 
of  IN  in  A.  Now,  by  applying  the  right-hand  rule,  we 
find  that  when  the  field  is  contracting,  there  is  a  current 
of  IN  induced  in  B,  which  is  in  the  same  direction  as  the 
dying  current  IN  in  A. 

If  we  again  look  at  the  end  view  of  the  coil  (Fig.  202) 
as  the  current  is  dying  out  in  A,  instead  of  increasing,  we 
see  that  the  field  due  to  the  dying  current  in  A  is  OUT 
and  the  field  due  to  the  induced  current  in  B  is  also  OUT. 


FIG.  201.— Effect  of  dying  field  in 
induction  coil. 


FIG.  202. — End  view  of  coil,  current 
in  A  dying. 


This  shows  that  the  field  due  to  the  induced  current  B 
opposes  the  decreasing  of  the  magnetic  field  of  current  in 
A  by  building  one  up  in  the  same  direction  as  the  dying 
field. 

This  is  as  we  should  expect,  since  we  have  seen  that 
inductance  is  analogous  to  inertia.  Inertia  reactions 
oppose  any  increase  or  decrease  in  the  existing  speed.  Induced 
currents  oppose  any  increase  or  decrease  in  the  existing  mag- 
netic field. 

161.  Direction  of  E.M.F.  Due  to  Inductance.  Lenz's 
Law.  THE  INDUCED  E.M.F.  TENDS  TO  SET  UP  A  CURRENT 
WHOSE  MAGNETIC  FIELD  ALWAYS  OPPOSES  ANY  CHANGE 
IN  THE  EXISTING  FIELD.  The  important  word  in  this  law 


276  ELEMENTS  OF  ELECTRICITY 

is  the  word  CHANGE.  The  field  of  the  induced  current  does 
not  always  oppose  the  existing  field,  nor  does  it  always  aid 
it,  but  it  always  opposes  any  change  in  it.  Thus  if  the  exist- 
ing field  is  zero,  and  we  send  a  current  through  the  pri- 
mary coil  to  set  up  a  field,  there  will  be  induced  in  the 
secondary  coil  a  current  whose  field  will  be  opposite  to  the 
growing  field,  and  thus  will  tend  to  keep  the  condition  of 
the  field  as  near  zero  as  possible.  But  if  there  is  a  field 
already  within  the  coil,  and  we  try  to  weaken  it,  say  by 
decreasing  the  current  in  the  primary,  there  will  be  induced 
in  the  secondary  coil  a  current  whose  field  will  aid  the 
existing  field  and  tend  to  keep  it  at  the  same  strength. 
The  induced  current  thus  is  always  such  that  its  field 
opposes  any  change  in  the  existing  magnetic  field.  This 
can  be  very  easily  proved  by  thrusting  a  magnet  into  a 
coil  of  wire  -which  has  a  galvanometer  in  the  circuit  and 
noting  that  there  is  a  current  induced  in  the  coil,  the  field 
of  which  opposes  the  direction  of  the  magnetic  field,  thus 
tending  to  neutralize  it  and  keep  the  condition  within 
the  coil  neutral.  When  the  magnet  is  withdrawn  there 
is  a  current  induced,  the  field  of  which  is  in  the  same  direc- 
tion as  the  magnet  field,  thus  tending  to  keep  up  the 
existing  condition  of  the  field. 

162.  Induction  Coils:  Jump  Spark.  One  type  of 
induction  coil  often  used  for  igniting  the  charge  in  a  gas 
engine  depends  upon  this  induced  current  in  the  secondary 
coil. 

The  primary  has  comparatively  few  turns  of  heavy  wire; 
the  secondary  has  many  turns  of  fine  wire.  A  core  of  soft 
iron  is  put  within  the  coils  to  strengthen  the  magnetic 
field.  See  Fig.  203.  The  secondary  circuit  B  has  an 
air  gap  in  it.  This  gap  is  between  two  points  (R  and  R\) 
which  are  inside  the  cylinder  of  the  gas  engine.  The 
primary  circuit  A  is  connected  to  a  set  of  battery  cells. 
The  cam  first  closes  the  primary  circuit  and  causes  a  field 
to  be  set  up  in  the  coil.  Then,  as  the  cam  opens  the  primary 


INDUCTANCE 


277 


as 


circuit,  the  field  is  suddenly  diminished  as  the  current  in 

the  primary  dies  out.     This  causes  a  high  induced  E.M.F. 

to  be  set  up  in  the  secondary,  because  the  field  on  dying 

out  cuts    the  large    number    of 

turns    in    the    secondary    at    a 

rapid  rate.     This  induced  E.M.F. 

is  high  enough  to  cause  a  spark 

to    jump    between    the     points 

R  and  RI,  and  ignite  the  charge 

of  gas  in   the    cylinder.      There 

is     a     small     spark    when    the 

circuit  in  the  primary  is  made, 

because    the   growing    field  cuts 

the     wires     of     the     secondary 

coil.     But  the  field  grows  much 

slower  than  it  dies  out  and  thus 

the    induced    E.M.F.    is   not 

great    and    the    spark    is 

more  feeble. 

163.  Ruhmkorff  Coils.  A  Ruhmkorff  coil  is  built  on 
the  same  plan  as  the  jump  spark  coil.  The  number 
of  turns  in  the  secondary  is  many  times  the  number  in 
the  primary,  and  thus  the  voltage  across  the  secondary,  is 
many  times  that  across  the  primary.  The  voltage  of  the 
secondary  should  be  in  about  the  same  ratio  to  the  voltage 
of  the  primary  as  the  number  of  turns  in  the  secondary  is  to 
the  number  of  turns  in  the  primary.  Of  course  what  is 
gained  in  voltage  is  lost  in  amperage. 

The  Ruhmkorff  coil  has  an  interrupter  in  the  primary, 
which  works  very  rapidly,  sometimes  as  high  as  200  times 
a  second.  This  will  cause  a  continuous  discharge  of  sparks 
to  pass  across  the  spark  gap  in  the  secondary  when  the 
current  is  broken.  The  current  for  the  primary  is  usually 
supplied  by  battery  cells.  This  type  of  coil  is  in  general 
use  with  a  gas  engine,  in  which  case,  when  the  cam  in 
Fig.  203  closes  the  primary  circuit,  the  interrupter  makes 


FIG.  203. — Jump  spark  coil. 


278  ELEMENTS  OF  ELECTRICITY 

and  breaks  it  very  rapidly,  causing  a  series  of  sparks  across 
the  gap  in  the  secondary. 

164.  Transformers.  Transformers  are  mutual  induction 
coils  used  on  alternating  current  lines  for  raising  or  lowering 
the  voltage.  Each  transformer  consists  of  a  primary  coil  and 
a  secondary  coil  wound  on  a  core  of  soft  iron  or  annealed 
steel. 

We  have  seen  that  it  is  much  more  efficient  to  transmit 
power  at  high  voltages.  Accordingly,  the  coil  of  few  turns  is 
connected  to  the  generator  and  the  coil  of  many  turns  to 
the  line.  The  voltage  of  the  line  is  then  approximately  as 
many  times  higher  than  the  voltage  of  the  generator  as  the 
number  of  turns  in  the  line  coil  is  greater  than  those  in  the 
coil  connected  to  the  generator.  A  transformer  so  connected 
is  called  a  "  STEP-UP  "  transformer.  Since  it  is  difficult 
to  use  this  high  voltage  for  driving  motors,  etc.,  it  is  "  stepped 
down  "  wherever  it  is  to  be  used.  This  is  done  by  con- 
necting in  the  same  kind  of  a  transformer,  putting  the  high 
voltage  side  (the  one  with  the  large  number  of  turns)  in 
the  line,  and  connecting  the  low  tension  side  to  the  motor. 
Thus,  by  means  of. two  transformers  of  the  same  kind, 
alternating  current  power  may  be  generated  and  used  at 
low  voltage,  yet  transmitted  at  high  voltage.  The  trans- 
formers of  the  present  date  being  remarkably  efficient, 
the  loss  in  transmission  is  thus  reduced  to  a  minimum. 

Transformers  are  made  in  two  general  types:  (a)  the 
CORE-TYPE,  Fig.  204,  in  which  the  coils  are  wound  around 
the  iron  cores.  (6)  The  SHELL-TYPE,  in  which  the  iron  core 
is  built  around  the  coils. 

Figs.  205  to  208  present  a  good  idea  of  the  appearance 
and  construction  of  a  shell  type  transformer.  Fig  205 
illustrates  the  appearance  of  the  transformer  in  its  iron 
case.  The  leads  or  "  taps  "  coming  through  the  case  are 
the  ends  of  the  low  voltage  coils.  Note,  by  inspecting  the 
diagrams,  that  both  the  primary  and  secondary  coils  are 
divided  into  two  parts.  This  gives  a  possibility  of  two 


INDUCTANCE 


279 


voltages  being  taken  from  the  transformer,  by  permitting 
the  coils  to  be  joined  either  in  series  or  parallel.     Note  also 


SECTION  A-A 


FlG.  204. — Core  type  of  trans- 
former. 


FIG.  205. — Fort  Wayne  transformer, 
shell  type, 


FIG.  206. — View  of  core  and  coils. 

that  the  core  is  laminated  to  reduce  the  eddy  currents  to 
a  minimum. 


280 


ELEMENTS  OF  ELECTRICITY 


.  207. — Another  view  of  core  and  coils. 


FIG.  208. — Arrangement  of  coils  in  two  sections. 


INDUCTANCE 


281 


A  transformer  is  usually  represented  by  a  diagram  as 
in  Fig.  209.  Sometimes  a  diagram  as  in  Fig.  210  is  used. 

The  action  of  a  transformer  follows  the  same  principle 
as  the  action  of  a  jump  spark  coil  except  that  the  current 


High  Tension  or 
Primary  Coils 


Low  Tension  or 
Secondary  Coil 


FIG.  209. — Conventional  diagram 
of  transformer. 


Secondary 
Coil 


Fi».  210. — Another  way  of  representing 
a  transformer. 


in  the  low  tension  side  instead  of  being  periodically  broken, 
is  periodically  reversed  by  being  connected  to  a  source  of 
alternating  current. 

The  current  in  the  high  tension  side  on  a  step-up  con- 
nection thus  opposes  not  only  the  dying  out  of  a  magnetic 
field  in  the  core,  but  also  an  actual  setting  up  of  a  field  in 
the  opposite  direction.  Since  current  in  the  low  tension 


0000000 


mm 


FIG.  211. — High  tension  transmission  by  means  of  transformers. 

coils  alternates,  the  field  in  the  core  continually  changes 
from  a  maximum  in  one  direction  to  a  maximum  in  the 
other,  and  there  is  thus  induced  in  the  high  tension  coils 
an  alternating  E.M.F.  which  tends  to  oppose  this  continu- 
ous change,  according  to  Lenz's  law. 


282 


ELEMENTS  OF  ELECTRICITY 


The  method  of  transmission  by  step-up  and  step-down 
transformers  is  shown  in  Fig.  211.  Notice  the  peculiar 
circumstance  that  the  alternating  current  motors  MI, 
M2  and  M3,  have  no  electrical  connection  with  the  generator 
G,  yet  they  draw  all  their  power  from  it.  Note  also  that 
the  transformers  TI,  T2  and  T3  are  connected  in  parallel. 

165.  Self  Induction.  Suppose  we  have  a  piece  of  wire 
1000  ft.  long  and  of  11  ohms  resistance.  If  we  stretch 
this  wire  out  straight,  "  line  and  return  "  for  transmitting 
power  500  ft.,  and  impress  110  volts  across  its  terminals, 
a  current  of  10  amperes  will  flow,  according  to  Ohm's  law. 


FIG.  212. — Induction  coil.    Self  induction. 


The  instant  we  close  the  switch  the  current  rises  to  a  steady 
value  of  10  amperes  and  remains  so  until  we  break  the 
circuit.  On  breaking  the  circuit,  a  very  slight  spark,  if 
any,  is  noticed. 

Now  suppose  we  wind  this  same  wire  into  a  single  coil 
on  an  iron  core  as  in  Fig.  212.  The  coil  still  has  11  ohms 
resistance.  When  a  pressure  of  110  volts  is  now  put  across 
its  terminals,  the  ammeter  in  series  with  the  coil  will  register 
10  amperes,  in  obedience  to  Ohm's  law.  But  it  will  not 
register  the  10  amperes  the  instant  that  the  switch  S  is 
thrown.  It  will  require  a  considerable  interval  of  time 


INDUCTANCE 


283 


to  acquire  the  steady  value  of  10  amperes.  If  we  should 
take  data  concerning  the  current  and  the  length  of  time 
since  the  switch  was  closed,  we  would  secure  some  such 
results  as  the  following: 


Time  Elapsed  since  Switch  was  Closed, 
in  Seconds. 

Current  in  Amperes. 

0 

0 

.1 

4.2 

.2 

6.6 

.3 

8.0 

.4 

9.0 

.5 

9.3 

.6 

9.6 

.7 

9.8 

.8 

9.9 

.9 

10.0 

1.0 

10.0 

Plotting   these   data,   for   current   strength   and   time,   we 
would  get  a  curve  as  in  Fig.  213. 


10 

8 

t« 

2 

2 



., 

, 

/ 

7 

/ 

/ 

.4  .6 

Time  in  Seconds 


1.0 


FIG.  213. — Relation  of  current  in  coil  to  time  elapsed  since  voltage  was  impressed 
across  terminals. 

Note  that  it  required  .9  second  for  the  current  to  rise 
to  its  full  value,  after  the  pressure  of  110  volts  was  thrown 
across  the  coil.  But  once  the  current  reached  the  value 
10  amperes,  which  it  should  have  according  to  Ohm's  law, 
it  remained  constant.  After  the  voltage  had  been  on  for 


284  ELEMENTS  OF  ELECTRICITY 

.2  second,  the  current  had  risen  to  only  6.6  amperes. 
There  must  be  some  property  in  a  single  coil  wound  in 
this  way  which  was  not  present  in  a  straight  wire  and  which 
now  opposes  any  rise  of  the  current. 

If,  as  before,  we  pull  the  switch  S,  we  find  that  the  current 
has  a  tendency  to  keep  flowing,  as  is  evidenced  by  the 
spark  at  the  breaking  points  of  the  switch.  There  is 
evidently  some  property  in  the  circuit  in  this  form  which 
opposes  the  dying  out  of  the  current.  This  property  in 
a  single  circuit  which  opposes  both  the  building  up  and 
the  dying  out  of  the  current  is  called  SELF  INDUCTANCE. 
A  momentary  E.M.F.  which  is  set  up  in  the  opposite  direc- 
tion to  the  change  that  is  taking  place  is  called  the 
INDUCED  E.M.F.  It  is  this  E.M.F.  induced  in  the  circuit 
which  keeps  the  current  down.  The  current  in  any  circuit 
is  always  proportional  to  the  algebraic  sum  of  the  impressed 
E.M.F.  and  induced  E.M.F.  When  a  circuit  possesses 
high  self  inductance,  which  we  shall  see  it  does  when  it 
possesses  ?  strong  magnetic  field,  this  induced  E.M.F.  is 
great  if  wo  try  to  suddenly  build  up  a  current,  or  destroy 
a  current  in  the  circuit. 

During  the  time  the  current  was  increasing,  the  induced 
E.M.F.  was  opposing  the  impressed  E.M.F.  and  the  dif- 
ference between  the  two  voltages  determined  the  current. 
Thus,  at  the  end  of  .2  second,  since  the  current  was  only 
6.6  amperes  and  the  impressed  voltage  was  110  volts,  there 
must  have  been  an  induced  E.M.F.  which  opposed.  the 
impressed  E.M.F.  and  cut  the  current  down.  For  a  growing 
current  we  might  express  this  in  the  form  of  an  equation: 

._Ex-e 
~~ 


where      i=  momentary  value  of   growing  current; 
e=         "      .  "       induced  E.M.F.; 

.R=resistance  of  circuit. 


INDUCTANCE  285 

The  induced  E.M.F.  at  the   end  of  .2  second  can  then  be 
found  as  follows: 

IW-e 
-II-' 
e  =  110-66; 
=44  volts. 

At  the  end  of  .4  second,  the  induced    E.M.F.  would  be 
much  less: 


11     ' 

e  =  110-99 
=  11  volts. 

By  the  curve  being  steeper  at  the  .2  second  point,  we  see 
that  the  current  is  changing  more  rapidly  then  than  at  .4 
second  point. 

The  induced  E.M.F.  is  then  greatest  when  the  current 
is  changing  at  the  greatest  rate,  and  dies  out  entirely  when 
the  current  becomes  steady. 

As  the  current  is  dying  out,  the  self  .inductance  of  the 
circuit  tends  to  keep  it  flowing,  by  setting  up  an  induced 
E.M.F.  to  cause  a  current  to  flow  in  the  same  direction. 
This  induced  E.M.F.  is  also  proportional  to  the  rate  of 
change  in  the  current.  And  since  a  current  has  a  tendency 
to  die  out  very  rapidly  as  the  circuit  is  broken,  the  induced 
E.M.F.  may  be  very  great  and  cause  a  heavy  momentary 
surge  of  current.  This  is  noted  in  the  arcing  at-  the  break, 
when  a  circuit  of  high  self  inductance  is  broken. 

Thus  we  have  seen  that  even  a  single  circuit  possesses 
the  property  of  inductance  if  it  has  a  strong  magnetic 
field.  The  inductance  sets  up  an  E.M.F.  which  opposes 
any  change  in  the  existing  current  of  the  circuit.  This 
property  is  called  SELF  INDUCTANCE,  when  the  momentary 
E.M.F.  is  induced  in  the  same  circuit  in  which  the  current 
change  takes  place.  It  is  exactly  like  the  induced  E.M.F. 


286 


ELEMENTS  OF  ELECTRICITY 


which  the  property  of  MUTUAL  INDUCTANCE  sets  up  in 
another  coil.  The  induced  E.M.F.  in  each  case  obeys 
Lenz's  law  and  tends  to  set  up  a  current  which  shall  oppose 
any  change  in  the  existing  magnetic  field. 

166.  Cause  of  Self  Inductance.  Make  and  Break  Spark 
Coil.  Consider  Fig.  214.  A  current  is  made  to  enter  at 
M.  It  first  reaches  the  wire  A\.  A  field  in  growing 
around  ^li  cuts  A3,  and  by  the  right-hand  rule,  induces  an 
E.M.F.  which  tends  to  send  a  current  OUT.  But  a  current 
of  IN  is  being  sent  through  A3.  This  growing  current  of 
IN  in  A3  is  thus  opposed  by  the  E.M.F.  of  OUT  induced  by 
the  growing  field  around  A\.  Similarly,  the  current  in 


A, 

FIG.  214. — Self  induction.    Effect  of  growing  current. 

A 5  is  opposed  by  an  E.M.F.  induced  by  the  growing  field 
around  ^.3,  and  the  current  in  A4  by  the  growing  field  of 
current  in  A^,  etc. 

Also,  if  we  consider  the  entire  field  within  the  coil,  we 
see  that  an  E.M.F.  is  induced  in  the  coil,  which  tends  to  set 
up  a  current  opposing  the  building  up  of  a  field  within  it. 
Thus  a  growing  current  in  such  a  coil  is  said  to  be  "choked  " 
back,  and  takes  some  measurable  length  of  time  to  come 
up  to  the  full  value  as  expressed  by  Ohm's  law. 

In  a  like  manner,  when  the  circuit  is  broken  and  the 
field  around  the  wires  and  within  the  coil  starts  to  die  out, 
the  wires  are  again  cut  by  the  field,  though  this  time  in  a 
direction  which  sets  up  an  E.M.F.  tending  to  maintain 
the  field  and  current  as  it  is. 


INDUCTANCE 


287 


0 


-  i 

i 


If  the  magnetic  field  is  sufficiently  strong  and  the  current 
is  broken  suddenly,  the  voltage  induced  by  the  dying  field 
cutting  the  wires  of  the  coil  may  be  many  times  the  original 
impressed  voltage.  Use  is  made  of  this  fact  in  the  MAKE 
AND  BREAK  SPARK  COIL. 

A  single  coil  of  many  turns,  as  in  Fig.  215,  is  wound  on 
core  of  soft  iron  wires.     A  moving  point  P2,  makes  contact 
with  a  stationary  point  PI.     Both  are  inside  the  cylinder 
of  a  gas  engine.  -  This  causes 
a  current  to  flow  through  the 
coil    and    set    up    a     strong 
magnetic    field.      When    the 
point  P2  leaves  PI  there  is  a 
sudden  breaking  down  of  this 
magnetic  field  which  sets  up 
enough    induced     E.M.F.    to 
cause  a  spark  to  jump  across 
the  gap  now  existing  between 
PI  and  P2. 

It  may  be  stated  as  a  general  principle  that  whenever 
an  inductive  circuit  (one  having  a  strong  magnetic  field) 
is  suddenly  broken,  the  induced  E.M.F.  is  likely  to  set  up 
an  arc  at  the  breaking  point. 

167.  Inductance  a  Property  of  the  Circuit.  The  prop- 
erty of  inductance  is  a  property  of  the  circuit,  not  of  the 
electric  current  or  voltage.  It  depends  entirely  upon  the 
shape  and  size  of  the  circuit  and  upon  the  magnetic  per- 
meability of  the  surrounding  medium.  If  the  medium 
consists  of  iron,  which  has  a  high  permeability,  and  the 
circuit  consists  of  a  coil  of  wire,  the  inductance  of  the  circuit 
is  great.  On  the  other  hand,  if  the  circuit  consists  of  a 
straight  wire  strung  in  the  air,  which  has  low  permeability, 
the  inductance  of  the  circuit  will  be  extremely  small.  The 
current,  voltage,  power,  etc.,  have  nothing  to  do  with  the 
amount  of  inductance  of  the  circuit,  though  these  are 
often  affected  by  the  inductance. 


FIG.  215. — Induction  coil.     Make  and 
break  spark  coil. 


288  ELEMENTS   OF  ELECTRICITY 

INDUCTANCE,  then,  is  not  a  material  thing,  but  merely 
a  term  which  expresses  the  result  of  a  certain  arrangement 
of  wires,  iron,  air,  etc.,  in  an  electric  circuit.  It  has  been 
likened  to  MOMENT  OF  INERTIA,  which  is  not  a  material 
thing,  but  merely  a  term  which  expresses  the  result  of  a 
certain  arrangement  of  weights,  etc.,  about  an  axis. 
Moment  of  inertia  is  not  a  property  of  matter,  force,  or 
energy,  but  of  the  machine  which  conveys  the  energy;  that 
is,  of  the  energy  circuit. 

And  just  as  the  moment  of  inertia  is  represented  by  an  alge- 
braic equation,  so  is  the  inductance  also.  In  fact,  INDUCT- 
ANCE can  be  defined  as  the  name  of  a  certain  algebraic  expres- 
sion relating  to  the  shape,  size,  etc.,  of  an  electric  circuit. 

Now  that  we  are  familiar  with  the  effect  of  this  property 
of  inductance,  we  may  study  the  algebraic  expression  by 
which  it  is  represented  and  measured. 

168.  Computation  of  Self  Inductance.  We  have  seen 
that  the  effect  of  inductance  in' an  electric  circuit  is  anal- 
ogous to  the  effect  of  inertia  in  a  moving  machine.  Inertia 
produces  a  reactive  force  which  opposes  any  change  of  speed 
in  the  machine.  The  amount  of  this  reaction  is  measured 
by  the  product  of  the  mass  of  the  machine  times  the  rate 
of  change  produced  in  the  velocity.  That  is,  we  write  the 
equation : 

/= mass  X  rate  of  change  of  speed, 
or 

/-ma, 

where  /= reacting  force  due  to  inertia; 

m=mass  of  body; 
a  =  acceleration  (rate  of  change  of  speed). 

If  we  use  "  v  "  (velocity)  instead  of  "a"  (acceleration), 
and  consider  the  body  to  start  from  rest  ("  0  "  velocity) 
and  to  get  up  "  v  "  ft.  per  sec.  in  "  t  "  seconds,  we  could 
express  the  "  rate  of  change  of  speed  "  by  the  quantity 

—  instead  of  by  a. 


INDUCTANCE  289 

We  might  then  write  the  equation  as, 


Similarly,  inductance  produces  a  reacting  force  (an  electro- 
motive force)  which  opposes  the  change  of  the  current, 
We  may  therefore  write  the  equation, 

E  (force)  =L  (inductance)  X  rate  of  change  of  current. 

If  we  consider  the  current  to  start  at  "  0  "  and  rise  to 
"  7  "  amperes  in  "  t  "  seconds,  we  can  represent  the  "  rate 

of  change  of  current  "  by  the  quantity  —  . 
Our  equation  then  becomes 

(1)  E=LX1' 

where       E  =  induced  electromotive  force  in  volts', 
L=  inductance  in  henrys; 
1=  current  in  amperes', 
£=time  in  seconds. 

A  HENRY  may  then  be  defined  as  the  Inductance  of  a 
circuit,  in  which  a  change  of  ONE  AMPERE  PER  SECOND 
produces  an  Induced  E.M.F.  of  one  VOLT. 

This   equation   is    similar   to    the    mechanics    equation, 

f=mX—,  in  which  we  might  define  the  unit  m  as  the  mass 
t 

of  a  body,  in  which  a  change  of  velocity  of  one  ft.  per  sec.  in  a 
second  produces  a  reaction  of  one  pound.  It  will  be  remem- 
bered that  we  do  not  measure  mass  as  a  quantity  in  itself, 
but  always  express  the  value  of  m  by  the  algebraic  quantity 

(  —  ),    which    contains  quantities  which  we   can   measure. 

\9/ 

In  the   same  way,  we   have   to  find  the  value   of    L   from 

an    algebraic    expression    containing   quantities   which   we 

can  measure. 


290  ELEMENTS  OF  ELECTRICITY 

This  algebraic  expression  for  the  value  of  L  for  a  coil 
can  be  worked  out  from  the  definition  and  from  the  above 
equation. 

We  have  seen  that  L  is  the  value  of  the  inductance  when 
E  volts  are  produced  by  a  change  of  /  amperes  in  t  seconds. 

We  know  that  one  volt  is  produced  by  one  wire  cutting 
10  8  lines  of  force  per  sec.,  or 

E 


t 

where       #=induced  E.M.F.  in  one  wire; 
</>  =  number  of  lines  cut; 
t  =time  of  cutting. 

Now  if  there  are  N  turns  of  wire  in  a  coil  which  cut  the 
lines,  then  the  E.M.F.  induced  in  the  coil  becomes: 


But  we   have   seen   by    Eq.    (1)   that   E=L'X-. 
Therefore  (1)=(2),  or 


t 
Multiplying  each  side  by  t 

m  u    ^ 

Ll    W 

We  have  seen  in  Chapter  VI  that 

«  *-£> 

and 

(5)  M  =  1.26#7, 
and 

(6)  «=J 


INDUCTANCE  291 

Then  by  substituting  (5)  and  (6)  in  (4)  we  get, 

(7)         |||  ^  =  1.26A^-4  || 

Substituting  this  value  of  <j)  in  Eq.  (3)  we  obtain, 


1OT 
and  dividing  through  by  /, 


,Q.        r       . 

(8)       L  =  -  ins/      '  sometimes  written 

where       L=  inductance  in  henry  s; 
N  =  number  of  turns; 
fji  =  permeability  of  core; 
A  =  section  area  of  core  in  sq.cms.-, 
Z=length  of  core  in  centimeters. 

This  is  the  usual  equation  which  is  used  whenever  it  is 
desired  to  find  the  inductance  of  a  given  coil,  wound  on 
a  given  core. 

If  the  coil  is  a  solenoid  with  mean  radius  r,  and  length 
I,  with  an  air  core,  the  equation  may  be  changed  as  follows: 


A  =^r2; 
/<  =  !. 
Thus  the  equation: 


(8)  L  =  - 
becomes, 

(9)  L  = 


1QPI 


Inasmuch  as  we  are  generally  dealing  with  the  induc- 
tance of  transformer  coils  and  electric  appliances  contain- 
ing more  or  less  iron  in  the  circuit,  Eq.  (8)  is  of  the 


292  ELEMENTS  OF  ELECTRICITY 

greatest    practical    importance.     The    two    may    be    used 
interchangeably  without  any  very  great  error. 

Example  (1).  What  is  the  inductance  of  the  primary  coil  of  a 
transformer  having  400  turns,  if  the  iron  core  is  60  centimeters 
long  and  300  sq.cms.  section  area?  Assume  j&t  = 


LJ  — 1 

AT  =  400  turns; 
A  =  300  sq.cms; 

1.26X400X400X1500X300 

108X60 
=  15.1  henrys. 

Example  (2).  Suppose  the  current  in  the  above  coil  to  change 
from  26  to  2  amperes  in  3  seconds,  what  average  voltage  would 
be  induced? 

I 


/     26-2 

—  =  —  -  —  =  8  amps,  per  sec.; 
t          o 

E  =  LXS 
=  15.1X8 
=  120.8  volts. 

Example  (3).  If  the  above  change  had  taken  place  in  .004  of 
a  second,  what  average  voltage  would  have  been  induced  in  the 
coil? 

/     26-2 

—  =  =  6000  amps,  per  sec.  ; 

t         •LHJ'A 


=  15.1X6000 
=  90,  600  volts. 

Problem  1-10.  A  coil  of  800  turns  is  wound  on  a  wrought 
iron  ring,  the  mean  diameter  of  which  is  15  cms.,  cross-section 
area  20  sq.cms.  Permeability  of  iron  =  1200.  What  is  the 
inductance  of  the  coil? 

Problem  2-10.  A  current  of  30  amperes  is  flowing  in  coil  of 
Problem  1.  If  the  current  is  reduced  to  18  amperes,  the  decrease 


INDUCTANCE  293 

taking  place  in  one-quarter  second,  what  average  voltage  will  be 
induced? 

Problem  3-10.  Compute  the  inductance  of  a  coil  75  cms.  long, 
4  cms.  in  diameter,  containing  2000  turns.  Air  core. 

Problem  4-10.  The  core  of  a  transformer  has  two  coils  wound 
on  it.  The  primary  consists  of  120  turns,  the  secondary  of  1200 
turns.  The  core  is  annealed  sheet  steel,  average  permeability 
of  2000,  72  cms.  long,  140  cms.  cross-section  area.  What  is 
the  inductance  of  primary  and  secondary  coils? 

Problem  5-10.  A  "  choke  "  coil  is  formed  by  winding  200 
turns  on  a  ring  consisting  of  annealed  sheet  iron,  permeability 
1800,  60  cms.  long  and  200  sq.cms.  cross-section  area.  What  is 
the  inductance  of  circuit? 

169.  Computation  of  Mutual  Inductance.  When  two  coils 
are  placed  together  in  such  a  way  that  an  increase  or  decrease  in  the 
current  in  the  turns  on  one  coil  causes  a  cutting  of  lines  of  force 
by  the  turns  of  the  other  coil,  they  are  said  to  have  a  MUTUAL 
INDUCTANCE,  as  we  have  seen.  The  symbol  for  mutual  inductance 
is  Lm,  and  is  equa]  to  unity,  or  one  henry,  when  a  change  of  one 
ampere  per  second  in  one  coil  causes  an  induced  E.M.F.  of  one  volt 
in  the  other. 

Jump  spark  induction  coils,  Ruhmkorff  coils  and  transformers 
depend  upon  this  principle  of  mutual  induction  for  their  action, 
as  explained  at  the  beginning  of  the  chapter. 

The  equation  for  the  value  of  the  mutual  inductance  of  two 
coils  becomes  very  simple  if  the  two  are  wound  on  the  same 
magnetic  core  and  have  about  the  same  diameter. 

_47r 


where      Nl  =  number  of  turns  on  one  coil  ; 

N2  =  number  of  turns  on  the  other  coil  ; 
r  =  radius  of  the  inner  coil  in  centimeters. 

Note  that  the  equation  is  the  same  as  that  for  the  self-inductance 
of  the  two  coils  wound  as  one  on  a  core,  except  that  instead  of 
squaring  the  total  number  of  turns  in  both  coils,  the  number  of 
turns  on  one  coil  is  multiplied  by  the  number  of  turns  on  the 
other. 

In  fact,  this  equation  for  mutual  inductance  can  easily  be  derived 
in  the  same  way  as  that  for  self-inductance  by  taking  into  account 
the  fact  that  the  changing  current  in  the  coil  with  NI  turns  causes 
the  conductors  of  the  coil  with  N2  turns  to  be  cut  by  the  changing 
flux,  and  induces  voltage  in  them. 


294  ELEMENTS  OF  ELECTRICITY 

Problem  6-10.  What  is  the  mutual  inductance  of  two  coils 
wound  on  a  circular  wooden  core  4  cms.  in  diameter  and  60  cms. 
long?  One  coil  contains  200  turns,  and  the  other  400  turns. 

Problem  7-10.  If  the  core  of  Problem  6-10  were  of  iron  (JJL  = 
1600),  what  would  the  mutual  inductance  be? 

Problem  8-10.  (a)  If  the  current  in  coil  containing  200  turns 
Problem  6-10,  were  changed  at  the  rate  of  100  amperes  per  sec., 
\vhat  voltage  would  be  induced  in  the  other  coil?  (6)  Answer 
same  question  for  coils  in  Problem  7-10. 

Problem  9-10.  If  the  current  in  coil  containing  400  turns, 
Problem  7-10,  be  changed  at  the  rate  of  100  amperes  per  sec., 
what  voltage  will  be  induced  in  the  other  coil? 

170.  Inductance  of  Transmission  Lines.  It  is  necessary  at 
times  to  compute  inductance  of  straight  parallel  conductors, 
especially  when  used  in  telephony,  telegraphy,  or  long  power 
transmission.  When  twin  non-magnetic  wires  are  strung  in  this 
way  in  the  air,  and  used  as  "line  and  return,"  the  inductance 
may  be  computed  by  means  of  the  following  equation : 


10" 

where  1= length  of  one  wire  in  centimeters; 

S  =  space  between  centers  of  wires  in  centimeters; 
r  =  radius  of  wire  in  centimeters. 

Problem  10-10.  Compute  the  inductance  in  a  100-mile  aerial 
line  consisting  of  two  copper  wires  20  inches  apart,  each  No.  6 
B.&S. 

Problem  11-10.  What  voltage  would  be  induced  across  the 
terminals  of  the  line  in  Problem  10-10  if  the  current  changed  from 
+  20  amperes  to  —20  amperes  in  -fa  of  a  second. 

171.  Effect  of  Self  Inductance  in  an  Alternating  Current 
Circuit.  Suppose  the  coil  in  Fig.  212  were  placed  in  an 
alternating  current  circuit  of  110  volts.  As  soon  as  the 
voltage  was  thrown  across  its  terminals  the  current  would 
start  to  rise  as  in  curve  213.  But  Jong  before  the  current 
had  time  to  reach  a  value  of  10  amperes,  the  voltage  would 


INDUCTANCE  295 

be  reversed  by  the  generator.  The  current  would  con- 
tinue to  rise  for  a  little  while,  although  the  voltage  was 
falling,  but  would  not  have  time  to  reach  the  value  of  10 
amperes  before  it,  too,  started  to  fall.  In  the  meantime, 
the  voltage  has  actually  reversed  and  is  now  tending  to 
force  a  current  through  the  coil  in  the  opposite  direction. 
Under  such  conditions  the  current  falls  very  rapidly,  and 
begins  to  flow  in  the  opposite  direction.  But  here,  too, 
long  before  it  reaches  a  value  of  10  amperes,  the  voltage 
that  started  it  flowing  in  this  direction  has  fallen  to  zero 
and  is  urging  it  to  flow  in  the  direction  in  which  it  started. 
Thus  the  effect  of  inductance  in  an  alternating  current 
circuit  is' to  keep  the  current  value  down,  or  to  "  choke  " 
it  back,  by  making  it  lag  so  far  behind  the  impressed 
voltage  that  it  does  not  have  time  to  get  up  to  the  value 
it  would  reach  on  a  direct  current  circuit.  The  higher  the 
frequency,  i.e.,  the  faster  the  voltage  is  reversed,  the  smaller 
the  current  in  the  circuit,  because  the  shorter  the  time 
in  which  it  has  to  grow  in  either  direction,  and  the  further 
it  lags  behind  the  voltage;  the  changes  in  current  taking 
place  considerably  later  than  the  corresponding  changes  in 
the  voltage. 


296  ELEMENTS  OF  ELECTRICITY 


SUMMARY  OF  CHAPTER  X 

The  kinetic  energy  of  an  electric  circuit  lies  in  its  mag- 
netic field.  The  strength  of  the  magnetic  field,  other  things 
being  equal,  depends  upon  the  shape  of  the  electric  circuit 
and  the  medium  in  which  the  field  lies.  It  is  convenient 
to  express  this  effect  of  "  shape  of  circuit  and  medium  of 
field  "  by  means  of  an  algebraic  equation. 

INDUCTANCE  is  the  term  applied  to  this  equation,  as  de- 
scribed above.  It  may  be  likened  to  the  algebraic  expression 
used  in  mechanics  and  known  by  the  term  Moment  of  Inertia. 
As  long  as  a  current  is  at  a  tonstant  value  in  a  circuit,  the 
property  of  Inductance  does  not  make  itself  known,  just  as 
the  moment  of  inertia  of  a  fly  wheel  does  not  affect  the  run- 
ning of  a  machine  as  long  as  the  speed  is  constant. 

LENZ'S  LAW.  By  means  of  this  property  of  inductance 
there  is  an  induced  E.M.F.  set  up,  whenever  the  strength  of 
any  magnetic  field  is  changed,  tending  to  cause  an  electric 
current  to  flow  in  such  a  direction  as  to  oppose  any  change 
taking  place  in  the  strength  of  the  field. 

SELF  INDUCTANCE  is  the  property  which  causes  this 
induced  E.M.F.  to  be  set  up  in  the  same  electric  circuit  in 
which  a  change  in  current,  and  consequently  a  change  in 
magnetic  field,  is  taking  place. 

MUTUAL  INDUCTANCE  is  the  property  which  causes  an 
induced  E.M.F.  to  be  set  up  in  a  second  circuit,  by  a  change 
of  current  in  the  first. 

HENRY;  UNIT  OF  INDUCTANCE.  When  a  change  of 
one  ampere  per  sec.  sets  up  an  induced  E.M.F.  of  one  volt, 
the  circuit  is  said  to  possess  an  Inductance  of  one  henry. 
Symbol  (L). 

EQUATIONS    FOR    INDUCTANCE. 

I.26N2M-4 

Self  Inductance,  L  = ~-, 

IO  L 

or 

47r2N2r2 


Mutual  Inductance, 


I09l 

47T2N,N2r2 


INDUCTANCE  297 

For  straight  transmission  lines 


\g.2i  Iog10-  +  i 
'J-L—  *-±. 


io9 

INDUCTION  COILS.  JUMP  SPARK:  TWO  COIL.  By 
suddenly  breaking  the  current  in  the  primary  coil,  which 
is  of  few  turns  wound  on  an  iron  core,  a  high  E.M.F.  is  induced 
by  mutual  Induction,  across  the  terminal  of  the  secondary 
coil  which  has  many  turns.  If  these  terminals  are  separated 
by  a  small  space,  a  spark  will  jump  across  this  gap  when  the 
primary  circuit  is  broken.  Accordingly,  these  secondary 
terminals  are  often  located  within  the  cylinder  of  an  explosion 
engine  for  the  purpose  of  firing  the  explosive  mixture,  by 
means  of  the  spark  thus  produced. 

RUHMKORFF;  TWO  COIL.  An  automatic  device  in 
an  arrangement  like  the  above,  opens  and  closes  the  primary 
circuit  very  rapidly,  thus  causing  a  flow  of  sparks  across  the 
terminals  of  the  secondary.  In  general  use  with  gas  engines. 

MAKE  AND  BREAK  SPARK:  SINGLE  COIL.  When 
even  a  small  current  is  broken  in  a  single  coil  of  many  turns 
wound  on  iron  core,  the  self  induction  causes  a  spark  to 
jump  across  the  break.  The  break  may  be  made  to  occur 
in  the  cylinder  of  a  gas  engine  and  explode  the  mixture. 

TRANSFORMERS  consist  of  primary  and  secondary  coils 
wound  on  a  ring  of  laminated  annealed  steel.  An  alternating 
current  in  one  coil  causes  a  continual  change  in  the  magnetic 
field.  This  induces  an  E.M.F.  in  the  other  coil.  The  ratio 
of  the  E.M.F.'s  across  the  two  coils  approximately  equals 
the  ratio  of  the  number  of  turns  in  the  two  coils.  Trans- 
formers can  therefore  be  used  to  raise  or  lower  the  voltage 
in  an  A.C.  circuit.  Efficiency,  very  high. 

EFFECT  OF  SELF  INDUCTANCE  IN  A.C.  CIRCUITS. 
(i)  Reduces  the  value  of  the  current.  (2)  Causes  current 
to  "lag"  behind  the  voltage. 


298  ELEMENTS  OF  ELECTRICITY 


PROBLEMS  ON  CHAPTER  X 

12-10.  The  current  in  a  circuit  is  changed  from  0  to  80  amperes 
in  2  seconds.  The  induced  voltage  is  16  volts.  What  is  the 
inductance  of  the  circuit? 

13-10.  What  voltage  is  required  to  reverse  a  current  of  80 
amperes  in  2  seconds  in  circuit  of  Problem  12-10? 

14-10.  How  often  per  second  will  an  alternating  E.M.F.  of 
220  volts  reverse  a  current  of  4  amperes  in  a  circuit  of  .05  henry 
inductance? 

15-10.  Find  the  inductance  of  a  coil  14  cms.  long,  8  cms. 
diameter,  having  200  turns  of  wire,  (a)  with  air  core;  (6)  with 
iron  core  of  2000  permeability. 

16-10.  Find  the  inductance  of  a  primary  coil  of  a  transformer 
having  1000  turns.  Length  of  magnetic  circuit  is  50  cms.,  area 
300  sq.cms.  Permeability  of  iron  =  1600. 

17-10.  If  secondary  coil  of  above  transformer  has  100  turns, 
what  is  the  inductance  of  its  circuit? 

18-10.  An  alternating  pressure  of  2000  volts,  maximum  value, 
which  alternated  120  times  (i.e.,  went,  from  maximum  value  in 
one  direction  to  maximum  value  in  the  other  direction  in  one  second) 
was  placed  across  the  primary  coil  of  transformer  in  Problem  16. 
What  was  the  maximum  current?  (Assume  zero  resistance  for 
primary  and  no  current  in  secondary). 

19-10.  How  much  voltage  would  be  induced  across  the  secondary 
coil  in  Problem  17,  if  the  conditions  of  Problems  16,  17,  and  18 
held  true? 

20-10.  An  iron  ring  (^  =  1000)  has  a  cross  section  of  40  sq.cms., 
and  an  average  length  of  90  cms.  How  many  turns  of  wire  must 
be  wound  on  it  to  produce  an  inductance  of  12  henrys? 

21-10.  When  500  turns  of  wire  are  wound  on  an  iron  ring,  20 
sq.cms.  section  and  average  length  of  75  cms.,  the  inductance  is 
1.2  henrys.  What  value  does  this  give  for  the  permeability  of 
iron? 

22-10.  If  1000  turns  were  wound  on  the  iron  ring  of  Problem  21, 
what  would  the  inductance  be? 

23-10.  A  transformer  core  of  annealed  steel  is  40  cms.  long 
and  120  sq.cms.  cross-section  area.  The  primary  coil  has  2000 


INDUCTANCE  299 

turns  of  wire  and  the  secondary,  200  turns.     Permeability  of  the 
core  is  2100.     What  is  the  mutual  inductance  of  the  coils? 

24-10.  What  is  the  self  inductance  of  each  of  the  two  coils  in 
Problem  23? 

25-10.  What  maximum  current  would  be  sent  through  primary 
coil  by  an  alternating  E.M.F.,  maximum  value  of  1100  volts, 
which  made  50  alternations  per  sec.?  Assume  zero  resistance 
and  no  current  in  secondary.  Use  data  of  Prob.  23-10. 

26-10.  What  voltage  would  be  induced  across  terminals  c 
secondary  in  Problem  25?  Compute  this  by  means  of  mutual 
inductance  and  rate  of  current  change  in  primary. 

27-10.  What  is  the  inductance  of  a  line  200  miles  long,  con- 
sisting of  No.  2  copper  hung  2.5  ft.  apart  in  the  air. 

28-10.  If  coils  on  core  in  Problem  23  are  joined  in  series,  what 
will  be  the  self  inductance  of  combination? 

29-10.  How  many  turns  of  wire  must  be  wound  on  core  in  Prob- 
lem 23  to  produce  2.8  henrys  inductance? 

30-10.  What  voltage  would  be  induced  in  coil  of  Problem  21 
by  a  current  changing  from  15  amperes  to  zero  in  ^^  sec.? 


CHAPTER   XI 
CAPACITY 

Capacity:  The  Elasticity  of  an  Electric  Circuit — Farad  and  Micro- 
farad, the  Unit  of  Capacity — Relation  of  Charge,  Voltage  and 
Capacity — Positive  and  Negative  Electricity — Bound  and  Free 
Charges — Condensers:  Equation  for  Capacity  of — Dielectric  Power 
— Dielectric*  Strength — Capacity  of  Cables;  Equation  for — Meas- 
urement of  Capacity;  by  Direct  Deflection  of  Ballistic  Galvano- 
meter; Bridge  Method — Locating  a  Break  in  a  Cable — Capacity 
of  Condensers  Joined  in  Parallel  and  in  Series. 

172.  Capacity;  Farad.  Inductance  has  been  likened 
to  inertia.  Capacity  may  be  likened  to  elasticity.  Thus 
if  the  walls  of  the  water  pipes  in  Fig.  309  were  elastic, 
either  along  the  whole  length  or  only  in  places,  they  would 
expand  with  each  forward  stroke  of  the  pump  and  contract 
with  each  back  stroke  and  thus  tend  to  equalize  the  pressure 
throughout  the  entire  stroke  by  acting  as  a  source  of  pres- 
sure. Similarly,  capacity  in  an  electric  circuit  gives  a 
certain  elasticity  to  the  circuit.  Should  the  impressed 
voltage  momentarily  fall,  if  there  is  considerable  capacity 
in  the  circuit,  it  will  tend  to  equalize  the  pressure  by 
acting  as  a  source  of  E.M.F. 

Inductance  makes  its  presence  known  when  the  current 
in  a  circuit  changes.  Capacity  makes  its  presence  known 
when  the  voltage  across  a  circuit  changes.  The  capacity  of  a 
condenser,  for  instance,  is  measured  by  the  amount  of 
electricity  that  can  be  added  to  it,  by  raising  the  voltage 
one  volt. 

It  is  like  the  capacity  of  a  tank  for  holding  compressed 
gas.  The  number  of  cubic  feet  of  gas  that  would  have  to 

300 


CAPACITY  301 

be  put  into  the  tank  to  raise  the  pressure  one  pound  per 
sq.in.  might  be  called  the  capacity  of  the  tank.  The 
capacity  of  the  tank  would  not  be  all  the  gas  it  could  hold 
without  bursting. 

In  the  same  way  the  number  of  coulombs  of  electricity 
that  must  be  added  to  a  condenser  to  raise  the  voltage  one 
volt  is  called  the  capacity  of  the  condenser.  The  capacity 
is  not  the  number  of  coulombs  it  can  hold  without  rupture. 

When  the  addition  of  ONE  COULOMB  raises  the  voltage 
across  the  condenser  ONE  VOLT,  we  say  that  the  condenser 
has  a  capacity  of  ONE  FARAD.  This  unit  is  so  large  that 
one-millionth  part  of  it  has  been  adopted  as  the  practical 
unit,  and  is  called  a  MICROFARAD. 

The  capacity  of  a  condenser,  then,  is  stated  in  terms  of 
the  quantity  it  will  hold  per  volt  pressure.  When  it  holds 
one  coulomb  for  every  volt  pressure  across  its  terminals, 
we  have  said  that  the  capacity  is  called  a  FARAD.  If  it 
holds  2  coulombs  per  volt  pressure,  its  capacity  is  2  farads, 
etc. 

The  capacity  being  stated  in  farads,  i.e.,  in  coulombs 
per  volt,  we  may  write  the  equation: 

r<fc      jx      Q  (coulombs) 
C(farad)  = 


where       C  =  capacity  in  farads; 
Q=  quantity  in  coulombs; 
E=  pressure  in  volts. 

Example  1.  How  many  coulombs  of  electricity  will  a  con- 
denser of  15  microfarads  capacity  hold,  when  the  pressure  between 
its  terminals  is  200  volts? 


-.000015X200 
=  .003  coulomb. 


302  ELEMENTS  OF  ELECTRICITY 

Example  2.     If  the  .003  coulomb  in  above  example  takes  ^ 
of  a  second  to  flow  into  the  condenser,  what  is  the  average  current? 


- 

~  t' 

_.OQ3 
~^002 
=  1.5  amperes. 

Problem  1-11.  What  is  the  capacity  of  a  condenser  that  holds 
.0012  coulomb  under  a  pressure  of  110  volts? 

Problem  2-11.  How  many  volts  would  be  required  to  put  .007 
coulomb  into  the  condenser  of  Problem  1? 

Problem  3-11.  The  capacity  of  a  condenser  is  20  mfs.  How 
many  coulombs  will  it  hold  when  the  pressure  is  220  volts? 

Problem  4-11.  What  is  the  average  charging  current  if  it 
takes  .08  sec.  for  condenser  in  Problem  3  to  become  fully  charged 
on  a  110-  volt  circuit? 

Problem  5-11.  If  condenser  in  Problem  4  be  placed  on  a  220- 
volt  circuit  and  it  took  same  length  of  time  to  charge,  what  would 
be  average  current? 

Problem  6-11.  If  condenser  in  Problem  4  be  charged  on  220-  volt 
circuit  and  then  disconnected  and  an  8-ohm  wire  be  placed  across 

its  terminals,  what  average  power  would  be  used  in  heating  the 

(<?*?n  \ 
Average  voltage  of  discharge  =-—  -.  ) 

Problem  7-11.  How  much  work  (in  watt-sec.)  would  be  done 
on  wire  in  Problem  6? 

Problem  8-11.  How  long  would  discharge  last  in  Problem  6? 
(Assume  average  rate  of  current.) 

Problem  9-11.  If  a  16-ohm  wire  were  used  in  Problem  6, 
what  would  be  average  current,  average  power,  and  total  work 
done  on  wire? 

Problem  10-11.  IT  a  16-ohm  wire  be  used  to  discharge  con- 
denser of  Problem  5,  what  would  be  average  current,  average 
power,  and  total  work  done  on  wire? 

Problem  11-11.  How  long  would  it  take  to  discharge  con- 
denser in  Problem  10?  (Assume  average  rate  of  discharge.) 


I 


CAPACITY  303 

Problem  12-11.  (a)  How  much  work  would  be  done  on  a  wire 
of  any  resistance  in  discharging  condenser  of  Problem  4?  (6)  In 
discharging  condenser  of  Problem  5? 

173.  Two    Kinds     of    Electricity.      According    to    the 
theory  which  best  explains  the  various  electrical  phenomena, 
there  are  two  kinds  of  electricity — a  positive  and  a  nega- 
tive.   Thus  a  battery  or  a  generator  sends  out  two  streams 
of  electricity,  one  flowing  around  the  circuit  in  one  direc- 
tion, the  other  in  the  other  direction.     In  Fig.  216,  the 
battery  cell  may  be  thought  of  as  sending  out  a  stream 
of  positive  charges  to  the  plate  A 
and    of    negative    charges    to  the 
plate  B,  since  A  is  connected  to  the 
positive  terminal    and    B    to    the 
negative  terminal  of  the  battery. 
According  to  this  theory  one^half    FlG  216._..Bound-  charges> 
of  the  work  in  an  electric    circuit 

is  done  by  the  positive  charges  florwing  in  one  direction, 
and  the  other  half  by  the  negative  charges  flowing  in  the 
other  direction.  This  part  of  the  theory  need  not  confuse 
us,  since  the  work  done  is  the  same  whether  it  is  all  done 
by  the  negative  or  all  by  the  positive  or  is  divided  between 
the  two. 

The  part  of  the  theory  that  particularly  interests  us 
is  that  which  deals  with  these  two  kinds  of  electricity, 
negative  and  positive,  in  their  effects  upon  one  another. 

It  has  been  found  by  experiment  that  the  laws  for  bodies 
so  charged  are  similar  to  the  laws  for  bodies  magnetized. 
It  must  not  be  inferred  from  this  statement  that  electricity 
and  magnetism  are  the  same.  As  has  been  stated,  they 
are  not  the  same,  though  possessing  peculiar  interrelations 
and  analogies. 

Electrified  bodies  resemble  magnetized  bodies  in  that 
bodies  charged  with  the  same  kind  of  electricity  repel  one 
another,  while  bodies  charged  with  unlike  kinds  attract 
one  another.  This  is  usually  illustrated  as  follows: 


304 


ELEMENTS  OF  ELECTRICITY 


If  we  bring  two  objects  near  together  which  have  unlike 
charges,  they  attract  each  other  with  considerable  force: 
and  the  charge  on  one  seems  to  attract  the  charge  on  the 
other. 

Thus  A  in  Fig.  217  represents  an  isolated  body  carrying 
a  positive  charge  of  electricity.  The  charge  is  practically 
uniformly  distributed  around  the  body.  If  a  body  B 
similarly  charged  be  brought  near  A,  the 
charges  would  appear  to  repel  one  another 
and  the  distribution  would  be  somewhat 
as  in  Fig.  218.  If,  however,  a  body  C 
negatively  charged  were  brought  near  A, 
the  charges  would  appear  to  attract  each 
other  and  the  distribution  would  be  as  in 
Fig.  219.  The  two  charges  in  this  case  are  said  to  bind 
each  other. 

Now  suppose  body  A  were  connected  to  a  source  of  supply 
of  positive  electricity  as,  for  example, to  the  positive  terminal 
of  a  battery  cell  or  a  generator,  and  body  C  were  connected 
to  a  source  of  supply  of  negative  electricity,  for  example, 
the  negative  terminal  of  a  cell  or  a  generator.  Fig.  216 
represents  such  a  case.  It  is  easy  to  imagine  now,  how 


FIG.  217.— Isolated 
charged  body. 


FIG.  2 18. —Distribution  of  charge  on 
two  bodies  carrying  like  charges. 


FHS.  219. — Two  bodies  carrying 
unlike  charges. 


the  positive  charge  on  A  draws  a  much  larger  negative 
charge  from  the  battery  cell  out  to  B  than  if  A  were  not 
near  B.  Also  the  negative  charge  on  B,  in  turn,  attracts 
a  much  larger  charge  of  positive  electricity  from  the  battery 
cell  out  to  A.  The  nearer  A  and  B  are  together,  the 
greater  this  attraction  will  be,  and  the  greater  the  charge 
each  will  have.  In  this  way  the  binding  effect  of  one  charge 


CAPACITY  305 

on  another  is  very  evident.  The  result  is  that  two  plates 
will  hold  a  larger  charge,  with  the  same  voltage  across  them, 
if  they  are  near  together,  than  if  they  are  separated  by 
a  large  distance.  They  will  thus  have  a  greater  capacity 
according  to  our  definition  of  capacity.  Of  course  the 
larger  the  plates,  the  greater  the  capacity  also. 

The  capacity  of  a  piece  of  electrical  apparatus,  then, 
depends  upon  the  area  of  parts  charged  with.  unlike  charges, 
and  upon  the  distance  between  them.  Experiments  show 
that  the  capacity  of  two  plates  such  as  A  and  B  in  Fig. 
216  is  directly  proportional  to  their  area  and  inversely 
proportional  to  the  distance  between  them.  This  may 
be  expressed  algebraically  as  follows: 


or 


where  C=  capacity; 

A  =  inside  area  of  plates; 
d=  distance  between  plates; 

K  =  constant  whose  value  is  determined  as  explained 
later. 

174.  The  Condenser.  According  to  the  above  pro- 
portion, if  we  wish  to  construct  a  piece  of  apparatus  with  a 
large  capacity,  we  must  place  large  plates  very  near  to  one 
another.  This  is  done  by  putting  thin  strips  of  mica  or 
oiled  paper  between  sheets  of  tin  or  lead  foil.  The  sheets 
of  foil  compose  the  plates  and  the  mica  acts  as  an  insulator. 
In  such  a  case  it  is  customary  to  call  the  mica  the  DIELEC- 
TRIC. A  piece  of  apparatus  so  constructed  is  called  a 
condenser.  Fig.  220  shows  the  conventional  way  of  repre- 
senting such  a  condenser.  Note  that  every  other  plate  is 
joined  to  one  side  of  the  line,  and  the  rest  of  the  plates  to 
the  other  side.  This  gives  large  negative  and  positive 


306  ELEMENTS  OF  ELECTRICITY 

plate  areas  which  are  very  near  each  other.  A  high  ca- 
pacity is  therefore  produced  by  the  BINDING  effect  of  the 
two  unlike  charges.  The  dielectric  is  not  represented. 

Such  a  condenser    placed  in  a  circuit  acts  like  an  air 
chamber  on  a  circuit  with  a  pump.     It  tends  to  counter- 
act the  effect  of  Inductance,  and  to  steady  the  voltage, 
just  as  an  air  chamber  counteracts 
the   effect    of   the   inertia    of    the 
water,  and  steadies  the  pressure  of 
water.     It  gives  a  kind  of  elasticity 
to  the  circuit,  acting  as  a  reservoir 

Fio.' ^.-Conventional  way  of    for    ™?    SUFPluS     Quantity    of     eleC- 

beapttlrynting  a  condenser  and  tricity.  For  this  reason,  it  has  a 
very  important  part  in  alternating 

current  circuits  and  telephone  lines,  where  the  voltage 
and  current  are  constantly  changing. 

In  constructing  a  condenser  several  points  have  to  be 
considered : 

(1)  To  have  the  capacity  high, 

(a)  The  plates  must  have  a  large  total  area; 
(6)  They  must  be  as  near  together  as  possible ; 
(c)  The  dielectric  must  have  the  power  of  conveying 

the  influence  of  the  charges  through  it,  in  order 

that  they  may  bind  each  other. 

(2)  To  have  the  unlike  plates  well  insulated  from  each 
other, 

(a)  The  dielectric  must  have  a  strength  such  that  the 
voltage  across  it  will  not  rupture  it  and  cause  a 
charge  to  pass  through  it  in  the  form  of  an  arc. 
The  term  "  dielectric  power  "  must  not  be  confused  with 
"  strength  of  the  dielectric." 

When  we  speak  of  the  "  strength  of  the  dielectric  "  we 
mean  merely  its  insulation  qualities  or  ability  to  resist 
rupture  or  leakage  of  current.  The  expression  "  Dielectric 
Power,"  however,  is  a  technical  term  which  needs  further 
explanation. 


CAPACITY 


307 


175.  Dielectric  Power.  A  condenser  built  up  of  certain 
size  plates,  and  of  a  certain  number  of  plates,  placed  a 
certain  distance  apart,  with  mica  sheets  as  the  dielectric, 
will  have  about  eight  times  the  capacity  of  one  built  with 
the  same  dimensions,  but  the  air  spaces  between  the  plates 
acting  as  the  dielectric.  Mica  seems  to  possess  the  POWER 
of  conveying  the  binding  influence  of  one  charge  on  the 
other  to  a  much  greater  extent  than  does  air. 

This  power  of  conveying  the  binding  effect  of  one  charge 
on  another  is  called  DIELECTRIC  POWER,  and  is  possessed 
to  greatly  varying  degrees  by  different  substances. 

In  equations  for  calculating  the  capacity  of  electrical 
apparatus,  this  dielectric  power  is  usually  designated  by 
the  letter  K.  Below  is  a  table  of  the  value  of  K  for 
substances  commonly  used  as  dielectrics. 

DIELECTRIC   POWER    OF    VARIOUS  SUBSTANCES 


Material. 

K 

Air,  at  ordinary  pressure;  standard  
Manila  paper 

1.0000 
1  50 

Paraffin    ...                        

1.68  to  2.30 

Beeswax 

1  86 

Paraffin   solid                              

1.9936  to  2  32 

Resin 

1  77  to  2  55 

Ebonite  .                                    

2.05  to  3.15 

India  rubber  pure 

2  22  to  2  496 

Gutta  percha.                              .  !  

2.45  to  4.20 

Shellac 

2  74  to  3  60 

Glass                                               

3.013  to  3.258 

Mica 

4  00  to  8 

Porcelain                                          

4.38 

Flint  glass  light 

6  85 

Flint  glass  double  extra  dense  

10.10 

INDUCTIVITY  and  SPECIFIC  INDUCTIVE  CAPACITY  are 
other  names  for  this  DIELECTRIC  POWER.  Air  is  generally 
taken  as  the  standard  for  Dielectric  Power  and  all  other 
substances  are  compared  with  it. 


308  ELEMENTS  OF  ELECTRICITY 

176.  Capacity    of    Plate    Condensers.     The   capacity   of 
a  condenser  constructed  as  described  above  may  be  com- 
puted from  the  general  equation: 

or 

885AK. 
1010d  ' 

where   C=  capacity  in  microfarads; 

A  =area    (one  side)    of    all    the    dielectric    actually 

between  plates  in  sq.cms.; 
d=  average  thickness  of  dielectric  in  cms.; 
K=  dielectric  power. 

Example.  What  is  the  capacity  of  a  condenser  which  has 
2000  plates?  Dielectric  consists  of  sheets  of  paraffined  paper, 
.005  cm.  thick.  The  part  of  each  sheet  actually  between  plates 
has  an  area  of  16X20  cms. 

885X2.1X16X20X2000 

1010X.005 
=  24  microfarads. 

Problem  13-11.  What  is  the  capacity  of  a  condenser  consisting 
of  200  plates  of  lead  foil?  The  dielectric  consists  of  10X15  cms. 
mica  sheets,  .1  mm.  in  thickness. 

Problem  14-11.  What  charge  will  condenser  in  Problem  13-11 
hold  under  a  pressure  of  220  volts? 

177.  Capacity  of  Telephone  and  Telegraph  Cables,  etc. 

One  of  the  greatest  hindrances  to  sending  telegraphic 
messages  across  the  ocean  is  the  large  capacity  of  the 
submarine  cable.  The  insulating  material  around  the 
wire  acts  as  a  dielectric  between  two  conductors,  the  wire 
and  the  water.  The  great  length  of  the  cable  gives  a 
sufficiently  large  area  to  the  dielectric  to  produce  a  very 
high  capacity.  In  transmitting  a  message,  then,  it  is 
necessary  to  alternately  fill  up  and  empty  a  piece  of  appa- 
ratus of  great  capacity.  The  time  necessary  for  these 


CAPACITY  309 

operations  puts  many  difficulties  in  the  way  of  efficient 
and  rapid  transmission. 

Telephone  cables  being  laid  in  pairs,  and  often  in  con- 
duits, also  possess  considerable  capacity,  which  would  make 
it  almost  impossible  to  operate  long  lines,  were  it  not  for 
the  possibility  of  "  loading  "  a  line  with  inductance  coils. 
As  has  been  explained,  CAPACITY  and  INDUCTANCE  have 
practically  the  opposite  effect  on  a  circuit;  inductance 
tending  to  choke  back  the  flow  of  electricity,  capacity 
tending  to  increase  the  flow.  Under  the  proper  condi- 
tions, one  may  be  made  to  neutralize  the  other.  Thus  the 
inductance  of  the  "  loading  "  coils  on  a  telephone  line, 
practically  neutralizes  the  capacity  effect  of  the  long  insu- 
lated cables. 

Sometimes  condensers  are  used  in  a  similar  way,  to 
counteract  the  effect  of  inductance.  Examples  of  this 
are  frequently  found  in  alternating  current  practice.  See 
Chapter  XV  for  a  discussion  of  inductance  and  capacity 
in  alternating  current  circuits. 

178.  Capacity  of  Cables.  The  capacity  of  submarine 
cables,  and  of  cables  laid  in  metal  sheaths  is  an  important 
factor  in  telegraphy  and  telephony. 

The  capacity  of  such  lines  may  be  computed  by  means 
of  the  same  equation  as  for  a  plate  condenser.  In  which 
case,  (d)  is  the  thickness  of  the  insulation  in  centimeters, 
and  A  is  average  area  in  sq.  cms.,  found 
as  follows: 

In  Fig.  221: 

Let 

D  =  outside  diameter  of  insulation. 
d=inside 
£>'=average       "  " 

Then, 


,v 

Lf   =  -  —  ^~~i  FIG.  221.  —  Cross-section  of 

<*  insulated  cable, 


310  ELEMENTS  OF  ELECTRICITY 

The  average  area  A  of  the  insulation  would  then  be  the 
area  of  a  cylinder  whose  diameter  =Df  and  whose  length, 
I,  equals  the  length  of  the  cable. 

7 
A  =nD'l     or     nl- 


A  more  common  equation,  however,  for  the  capacity  of  a 
cable  is  the  following  : 

2A13KI 


where  C=  capacity  of  cable  in  microfarads; 

Z=length  of  cable  in  centimeters; 
D=  outside  diameter  of  dielectric  (insulation)  in  cms.; 
d=inside  diameter  of  dielectric  in  cms.; 
X=dieleotric  power  of  insulation. 

Problem  15-11.  A  No.  6  (B.  &  S.)  copper  wire  covered  with 
.05  inch  of  gutta  percha  insulation  is  held  in  a  lead  sheath.  Find 
capacity  per  mile  of  this  cable  and  grounded  sheath. 

Problem  16-11.  A  submarine  telegraph  cable  is  of  copper 
.4  inch  in  diameter,  and  covered  with  .6  inch  of  gutta  percha 
insulation.  What  is  capacity  between  cable  and  water  of  50 
miles  of  such  cable? 

THE  CAPACITY  OF  AN  AERIAL  LINE  of  twin  wires  may 
be  found  by  means  of  the  following  equation: 

12.51 


where   Cy=  capacity  in  microfarads; 

I  =  length  of  one  wire  in  centimeters; 
S=  space  between  wires  in  centimeters; 
r=radius  of  each  wire  in  centimeters. 


CAPACITY  311 

Problem  17-11.  What  is  the  capacity  per  mile  of  a  line  con- 
sisting of  two  wires,  .16  inch  in  diameter  and  separated  by  a  space 
of  20  inches? 

Problem  18-11.  What  is  capacity  of  a  20-mile  line  consisting 
of  two  No.  4  (B.  &  S.)  copper  wires  hung  18  inches  apart? 

179.  Measurement  of  Capacity.  Direct  Deflection  of 
Ballistic  Galvanometer.  In  measuring  capacity  we  make 
use  of  the  fact,  that  in  order  to  charge  a  circuit  with  elec- 
tricity, the  charge  must  flow  from  the  source  along  a  con- 
ductor. If  a  ballistic  galvanometer  is  inserted  in  series 
in  the  line,  this  flow  of  the  charge  can  be  made  to  give 
a  momentary  deflection,  or  THROW,  proportional  to.  the 
amount  of  the  charge. 

Thus  in  Fig.  222,  when  the  switch  S  is  closed,  a  positive 
charge  flowing  along  the  upper  conductor  to  charge  plate 
A}  must  pass  through  galvanometer  G.  This  will  cause 


_L 

T 


FIG.  222. — Ballistic  method  of  measuring  charge  flowing  into  condenser. 

the  galvanometer  to  "  throw."  Since  the  galvanometer 
is  constructed  so  that  the  needle  swings,  or  "  throws/'  to  a 
reading  proportional  to  the  moving  force  and  immediately 
returns  to  zero,  it  is  said  to  be  a  BALLISTIC  galvanometer. 
This  term  distinguishes  it  from  one  which  requires  a  steady 
current  to  cause  its  deflections  to  be  proportional  to  the 
moving  force. 

A  more  common  way  of  connecting  the  galvanometer 
is  shown  in  Fig.  223.  The  switch  S  is  thrown  down  to 
charge  the  condenser  plates  A  and  B,  and  then  quickly 
thrown  up  and  allowed  to  discharge  through  the  galvanom- 


312  ELEMENTS  OF  ELECTRICITY 

eter.      A  special  charge  and    discharge    switch  is  required 
for  accurate  results. 

To  measure  the  capacity  of  any  device,  such  as  a  given 
length  of  telephone  cable,  it  is  first  charged  for  one  minute, 
then  suddenly  discharged  through  the  galvanometer  and 
the  throw  noted.  A  condenser  of  known  capacity  is  then 
inserted  in  the  place  of  the  unknown  and  charged  for  a 

minute,  and  the  throw  of 
the  galvanometer  noted  as 
it  is  discharged.  The  ca- 
pacities of  the  two  pieces 
are  then  in  the  same  ratio  as 
the  galvanometer  throws, 
providing  the  throws  have 


1 1   1 1 

J|   IjJ      not  too    great   a  difference. 

Since  the  known  capacity 
is     usually     the    variable 


FIG.  223.— More  precise  ballistic  method     standard  condenser,  shown 

for  measuring  charge  in  condensers. 

in   Figs.  231    and   232,   its 

capacity  can  be  adjusted  until  the  galvanometer  throw 
on  discharge  is  very  nearly  equal  to  the  throw  caused  by 
the  unknown  capacity.  In  this  way  a  considerable  degree 
of  precision  can  be  obtained. 

Problem  19-11.  In  a  circuit  arranged  as  per  Fig.  223,  a  cable 
is  charged  to  a  given  potential.  On  being  discharged  through 
a  ballistic  galvanometer,  it  causes  the  instrument  to  deflect  12.6 
scale  divisions.  A  standard  condenser  of  .5  mfs.  capacity  after 
being  charged  to  the  same  potential,  on  discharge  causes  the 
same  galvanometer  to  deflect  11.8  scale  divisions.  What  is 
capacity  of  cable  as  tested? 

180.  Capacity  Measurement.  Bridge  Method.  Another 
method  for  measuring  capacity  is  shown  in  Fig.  224.  It 
resembles  a  Wheat  stone  bridge,  with  the  exception  that 
condensers  Ci  and  C2,  one  known  and  the  other  unknown, 
replace  two  of  the  resistances.  A  source  of  alternating 
current  supply  G  takes  the  place  of  a  battery  cell,  and  a 


CAPACITY 


313 


telephone  receiver  T  replaces  the  galvanometer.  G  is 
usually  a  small  hand  magneto.  A  balance  is  obtained, 
as  in  a  Wheatstone  bridge,  by  adjusting  RI  and  R2  until 
there  is  no  sound  in  receiver  T  when  the  magneto  G  is 
running. 

The   following   equation   is   then   used   to   compute   the 
capacity  of  the  unknown,  say  C2: 


R2    Ci 

Note    that   the   ratio   of  'the   capacities   is   the   inverted 
ratio  of  the  resistances.     In  this  respect  it  differs  radically 
B 


FIG.  224.— Bridge  Method  of 
measuring  capacity. 


FIG.  225. 


from  the  Wheatstone  bridge  equation  for  resistance  meas- 
urements. 

The  proof  of  the  above  equation  is  as  follows: 

When  there  is  no  sound  in  receiver  T  (Fig.  225),  B  and  E  must 
be  at  the  same  potential.  The  voltage  from  A  to  B  (IiRi)  must 
equal  the  voltage  from  A  to  E  (/2/22),  or 

(1)  hR,=hR,. 

Also  the  voltage  from  B  to  D  must  equal  the  voltage  from  E 
to  D.  The  voltage  across  a  condenser  equals  the  charge  in  it 
divided  by  its  capacity.  (See  paragraph  172.) 

If  qi  be  the  charge  on  Clf  and  qt  the  charge  on  Cz  at  any  instant, 
then 

Voltage  (B  to  />)=£; 
v  i 

Voltage  (tf  to  £)=;£•; 


314  ELEMENTS  OF  ELECTRICITY 

Therefore, 


or, 

(2)  Crft-Crf,. 

But  in  order  that  the  quantity  (<?,)  get  to  the  condenser  (d)  it 
must  flow  through  (#1)  as  there  is  no  flow  along  EB. 

In  like  manner  the  quantity  g2  must  flow  through  R^. 
Thus  we  have  equation, 

(3o)  A  (current  in  ^0  =  % 

and 

(36)  72  (current  in  /22)=y; 

where  t  =  the  time  in  seconds  in  which  the  charges  flow  through 
the  resistances.     This  time  must  be  the  same  in  each  resistance 
since  the  two  resistances  are  in  parallel  across  the  same  alternator. 
Then,  multiplying  (3a)  by  Rl  and  (36)  by  R2,  we  have, 


and 

#2 

t    2t 
substituting  these  values  in   (1), 

-V"  .*.•' 

or 

Dividing  equation  (4)  by  (2)  we  have, 


or 


= 

C,     Ct' 

This  may  be  transposed  to  read 


CAPACITY 


315 


181.  Locating    a    Break    in    a    Telephone    Cable.    The 

above  method  is  often  used  in  locating  a  break  in  one  of 
a  pair  of  telephone  cables.  In  Fig.  226,  Ci  represents  a 
pair  of  good  wires,  C2  represents  a  pair  in  which  one  of  the 
wires  is  broken  at  F. 


When  bridge  is  balanced,  jr^Tr 

the  capacity  of  the  faulty  pair. 
Now,  if  the  same  conditions 
exist  with  regard  to  the  wires 
(except  that  one  wire  of  one  pair 
is  broken)  the  capacity  of  the 
faulty  pair  as  far  as  the  break 
will  have  the  same  relation  to 
the  capacity  of  the  good  pair, 


where  C2  represents 


FIG.  226.— Capacity  test  for 
locating  break  in  cable. 


that  the  distance  out  to  the  break  has  to  the  length  of 
good  pair. 


Thus: 


where 


= 

h  cv 

/2=  distance  to  break; 
/i  =  length  of  good  pair; 

2  =  capacity  to  break ; 

i  =  capacity  of  good  pair. 


Therefore 


R 


If  the  length  of  the  good  pair  is  known,  the  distance  12 
out  to  the  break  can  easily  be  computed. 

Problem  22-11.  In  a  cable  test  for  a  break,  arranged  as  in 
Fig.  226,  the  length  of  the  pair  of  good  wires  (d)  is  2500  ft.  A 
balance  is  obtained  when  R1  =  14  ohms  and  #2  =  30  ohms.  How 
far  out  is  the  break  in  the  faulty  cable  (C2)? 


316 


ELEMENTS  OF  ELECTRICITY 


Problem  23-11.  If  the  capacity  of  the  good  pair  in  Problem 
22  is  2.83  mfs.,  what  is  the  capacity  of  the  faulty  pair  as  far  out 
as  the  break? 

182.  Condensers  in  Parallel.  Consider  two  condensers 
I  and  II,  Fig.  227,  joined  in  parallel  across  the  mains,  the 
voltage  of  which  is  E.  The  capacity  of  condenser  I  is 
Ci;  of  condenser  II,  C2.  Find  the  combined  capacity  of 
the  two  when  so  joined. 

Let   C  be  the  combined  capacity; 
"      Qi  =  quantity  of  charge  in  condenser  I; 
"      Q2=         "  "  "          II. 

Then  total  quantity  in  both  condensers  =  Qi  +Q2. 

(1)  Qi+Q2=CE. 


FIG.  227. — Condensers  in  parallel. 

(Quantity  of  total  charge  equals  total  voltage  times  total 
capacity.) 

But  QI=C!# 

and 

Q2=C2E. 

Therefore    (2)  Q}  +Q2  =  (Ci  +  C2)E. 
From  (1)  and  (2)  we  have 


or 


E  =  (Cl+C2)Et 
C  =  Ci+C2. 


CAPACITY 


317 


Thus  the  capacity  of  condensers  joined  in  parallel  equals 
the  sum  of  the  capacities  of  the  separate  condensers.  Join- 
ing condensers  in  parallel  is  merely  adding  the  plate  area 
of  one  to  that  of  the  other. 

Example.  What  is  the  capacity  of  4  condensers  of  3,  .2,  7, 
and  2.5  microfarads  respectively,  when  joined  in  parallel? 

C  =  C1-fC2+ 

C  =  3  4-  .2  +  7  +  2.5  =  12.7  microfarads. 

Problem  24-11.  What  charge  is  sent  into  condenser  I,  Fig. 
227,  when  #  =  110  volts?  Capacity  of  1  =  6  mfs. 

Problem  25-11.  (a)  What  charge  is  sent  into  condenser  II, 
Fig.  227,  when  #  =  110  volts?  (6)  What  charge  goes  to  combi- 
nation? Capacity  of  11  =  4  mfs. 

183.  Condensers  in  Series.  Condensers  in  series  present 
a  peculiar  phenomenon.  Let  condenser  I  of  capacity  Ci, 
Fig.  228,  be  joined  in  series  with  condenser  II  of  capacity 


FIG.  228. — Condensers  in  series. 

€2.     Let  E  be  voltage  across  combination,  E\  across  con- 
denser I,  and  E2  across  condenser  II. 

The  charge  Q  is  sent  into  the  condensers  under  the  action 
of  the  voltage  E.  Since  the  two  condensers  are  in  series 
the  same  charge  must  be  sent  into  each,  just  as  the  same 
current  is  sent  through  resistances  in  series.  Thus  the 
charge  sent  into  each  is  Q,  and  the  charge  sent  into  the 
combination  is  also  the  same  Q. 


318 


ELEMENTS  OF  ELECTRICITY 


FIG.  229. — Diagram  of  connections  for  a  starting  box  with  "no- voltage"  release. 


FIG.  230. — Diagram  of  connections  to  starting  box  with  "no-voltage"  arid  "over- 
load" release. 


CAPACITY  319 

Let  C  =  combined  capacity  of  C\  and  C<2. 


But 

(2)  E= 

Therefore,  from  (1)  and  (2), 

Q_Q_,Q_. 

C    Ci     CV 
and 

L-i+A 

C    d  +CV 

Thus  the  reciprocal  of  the  combined  capacity  of  con- 
densers in  series  equals  the  sum  of  the  reciprocals  of  the 
capacities  of  the  separate  condensers. 

Example.  If  the  capacity  of  Condenser  I  in  Fig.  228  is  2  mfs. 
and  that  of  condenser  II,  5  mfs.,  what  is  the  capacity  of  the  two 
condensers  joined  in  series? 

1__L+J_. 

c  c,    cy 


== 

C     25     10' 
C--  1.43  mfs. 


Problem  26-11.  What  charge  is  sent  into  condensers,  of  above 
example,  if  the  voltage  across  the  combination  is  110  volts? 

Problem  27-11.  What  is  the  voltage  across  condenser  I  and 
across  condenser  II,  Fig.  228,  if  voltage  across  the  two  in  series  is 
110  volte?  Capacity  of  1  =  8  mfs.,  of  11  =  3  mfs. 


320  ELEMENTS  OF  ELECTRICITY 


SUMMARY  OF  CHAPTER  XI 

CAPACITY:  The  property  which  gives  an  electric  circuit 
elasticity  and  tends  to  steady  the  pressure. 

FARAD,  UNIT  OF  CAPACITY:  A  piece  of  apparatus 
has  a  farad  capacity  when  it  holds  one  coulomb  of  electricity 
for  every  volt  pressure  across  its  terminals.  A  microfarad, 
the  practical  unit,  is  one-millionth  of  a  farad. 

THE  RELATION  BETWEEN  CHARGE,  VOLTAGE  AND 
CAPACITY  is  expressed  by  the  equation 


BOUND  CHARGES.  The  two  kinds  of  electricity,  positive 
and  negative,  attract  each  other,  like  two  unlike  magnetic 
poles.  Two  oppositely  charged  plates  near  each  other  have 
a  greater  capacity  on  account  of  this  attraction  than  when 
widely  separated.  The  charges  are  then  said  to  be  bound. 

PLATE  CONDENSERS.  EQUATION  FOR  CAPACITY  OF: 
Consist  of  thin  sheets  of  lead  or  tin  foil,  separated  by  thin 
sheets  of  insulating  material  called  the  dielectric.  The  large 
number  of  the  metal  plates  affords  a  large  area  on  which  to 
collect  the  charge,  and  the  oppositely  charged  plates  being 
separated  by  the  thickness  of  the  dielectric  only,  bind  the 
charges  on  one  another. 

The  equation  for  the  capacity  of  C  condenser  is 

886KA 

*  io10d  ' 

The  value  of  K  in  the  equation  depends  upon  the  material 
of  the  dielectric  and  is  called  its  DIELECTRIC  POWER.  It 
represents  the  power  of  the  dielectric,  as  compared  with  air, 
to  convey  the  binding  effect  of  one  charge  on  the  other. 

CAPACITY  OF  CABLES.  All  cables  have  some  capacity. 
Submarine  cables  have  a  large  capacity  due  to  water  forming 
one  plate,  separated  but  a  small  distance  by  the  insulation 
from  the  cable,  which  forms  the  other  plate. 

Usual  equation  for  capacity  of  a  submarine  cable  is: 


10' 


CAPACITY 

of  an  aerial  line- 
ns/ 


—  Xio8 

Capacity  may  be  measured:  (i)  By  direct  deflection  method, 
using  standard  condenser.  Deflections  of  ballistic  galvanom- 
eter on  discharge  of  standard  and  unknown,  charged  to 
same  potential,  are  proportional  to  the  respective  capacities. 
(2)  By  bridge,  using  source  of  A.C.  power,  standard  variable 
resistances,  standard  condenser,  and  telephone  receiver. 
Equation  of  balanced  bridge  is 


= 

R2~d' 

which  is  not  the  Wheatstone  bridge  equation. 

LOCATING  BREAK  IN  CABLE.  Apparatus  arranged  as 
in  bridge  measurement  of  capacity;  using  good  pair  of  twin 
wires  of  known  length  exactly  likje  the  faulty  pair.  Equation 
is 

li     Ra 

12  Rl" 

CAPACITY  OF  CONDENSERS  IN  PARALLEL.  Equals 
sum  of  their  separate  capacities. 

CAPACITY  OF  CONDENSERS  IN  SERIES.  Equals  the 
reciprocal  of  the  sum  of  the  reciprocals  of  their  separate 
capacities. 


PROBLEMS  ON  CHAPTER   XI 

28-11.  What  is  the  capacity  of  a  condenser  made  up  of  500 
plates  of  lead  foil  with  a  dielectric  of  paraffined  paper  10X9  cms. 
and  .003  cm.  thick? 

29-11.  What  is  the  capacity  of  a  condenser  having  500  plates 
of  lead  foil,  if  the  dielectric  consists  of  mica  sheets  10X9  cms. 
.003  cm.  thick? 

30-11.  If  condensers  of  Problem  28  and  Problem  29  were 
joined  in  parallel,  what  would  be  their  joint  capacity? 


322  ELEMENTS  OF  ELECTRICITY 

31-11.  If  condensers  of  Problems  28  and  29  were  joined  in 
series  what  would  their  joint  capacity  be? 

32-11.  What  charge  would  condensers  joined  as  in  Problem  30 
hold,  if  a  voltage  of  110  volts  were  placed  across  the  terminals? 

33-11.  If  the  voltage  across  the  condensers  connected  as  in 
Problem  31,  were  changed  from  115  to  15  volts,  in  .05  sec., 
what  average  current  would  flow  from  the  condensers? 

34-11.  In  a  testing  outfit  arranged  as  in  Fig.  226,  the  telephone 
receiver  gives  no  sound,  when  RI  =  40  ohms  and  R2  =  76  ohms.  If 
the  length  of  the  good  pair  of  cables,  Ct  is  4940  ft.,  how  far  out 
is  the  break  in  the  faulted  pair,  (72? 


FIG.  231. — Standard  condenser. 

35-11.  It  is  desired  to  build  up  a  capacity  of  2  microfarads. 
There  are,  at  hand  3  condensers  of  4,  1,  and  3  mfs.  respectively. 
Show  by  diagram  what  arrangement  of  these  condensers  will 
produce  a  capacity  nearest  to  2  mfs.?  Compute  exact  capacity 
of  such  an  arrangement. 

36-11.  A  condenser  of  4.2  mfs.  capacity  charged  at  110  volts 
is  suddenly  discharged  through  a  short  wire.  What  amount  of 
energy  in  watt-sees,  is  used  in  heating  the  wire?  Note. — Average 

E 
voltage  on  discharge  =—. 

37-11.  If  wire  in  Problem  36  has  a  resistance  of  5  ohms,  in 
what  time  will  condenser  be  discharged  through  it  at  average 
rate? 

38-11.  If  70  volts  are  used  to  send  messages  through  cable  in 
Problem  16,  how  long  does  it  take  for  each  charge  and  discharge 
of  cable  at  average  rate?  Assmne  return  of  zero  resistance. 


CAPACITY  323 

39-11.  If  the  same  size  wire  and  same  voltage  as  in  Problem 
16  had  been  used  in  aerial  50-mile  line,  how  long  would  each 
charge  and  discharge  take  at  average  rate?  Line  to  consist  of 
two  wires  hung  18  in.  apart. 

40-11.  A  10  microfarad  condenser  is  charged  to  a  potential 
of  110  volts  and  a  15-microfarad  condenser  to  a  potential  of  220 
volts.  They  are  then  joined  together,  positive  to  positive  and 
negative  to  negative.  What  is  voltage  across  the  parallel  com- 
bination? 


FIG.  232. — Diagram  of  electrical  eonnections  in  a  standard  condenser. 

41-11.  If  the  condensers  in  Problem  40  had  been  joined  in 
series  after  being  charged,  what  would  voltage  across  the  combina- 
tion become? 

42-11.  What  charge  would  there  be  on  each  condenser  in  Prob- 
lem 40  after  being  joined  in  parallel? 

43-11.  What  charge  would  there  be  on  condensers  in  Prob- 
lem 40  after  being  joined  in  series? 


CHAPTER  XII 
ELECTROCHEMISTRY 

Primary  and  Secondary  Cells — E.M.F.  Dependent  on  Materials — 
Chemical  Action  of  Simple  Cell — Electrolysis — Electrochemical 
Equivalents — Electroplating  and  Electrotyping — Storage  Cells; 
Theory,  Construction  of  Plante,  Faure,  and  Edison  Types — Use 
and  Maintenance — Advantages  and  Disadvantages. 

184.  Electromagnetic   Induction.     In    Chapter    VII.,    it 
has    been    shown  that  whenever    an    electrical    conductor 
cuts  magnetic  lines  of  force,  an  electromotive  force  is  set 
up  between  the  ends  of  the  conductor.     A  high  electro- 
motive force  is  secured  by  joining  a  large  number  of  these 
conductors  in  series   (in  an  armature)   and  causing  them 
to  cut  rapidly  a  field  of  great  intensity.     Such  a  machine, 
as  described  in  Chapter  VII.,  is  the    ordinary    generator. 
It  may  be  said  to  transform  the  mechanical  energy  supplied 
to  the  pulley  into  electrical  energy,  which  it  delivers  from 
the    armature    terminals.     This   is   the    usual    method    of 
obtaining  electric  pressure  for  power  purposes. 

185.  Electric    Battery.     Primary    and    Secondary    Cells. 
But  it  is  possible  to  transform  not  only  mechanical  energy 
but  also  chemical  energy  into  electrical  energy.     A  device 
for  developing  electric  pressure  by  means  of  chemical  action 
is  called  an  ELECTRIC  BATTERY. 

It  has  been  found  by  experiment  that  when  two  electrical 
conductors  of  different  materials  are  placed  in  a  liquid 
so  that  they  do  not  touch  each  other,  and  the  liquid  acts 
on  one  of  them  chemically,  then  an  electric  pressure  is  set 
up  between  the  conductors.  The  liquid  is  called  an  ELEC- 
TROLYTE, 

324 


ELECTROCHEMISTRY  325 

The  conductor  of  the  highest  potential  is  called  the 
POSITIVE  PLATE;  the  other,  the  NEGATIVE  PLATE.  If 
the  two  plates  are  then  joined  by  a  wire,  an  electric  current 
will  flow  through  it  from  the  positive  to  the  negative  plate. 
The  voltage,  or  pressure,  between  the  plates  depends 
entirely  upon  what  the  electrolyte  is,  and  upon  what  the 
materials  of  the  two  plates  are.  The  size,  distance  apart, 
etc.,  have  nothing  to  do  with  the  voltage  of  a  battery, 
any  more  than  the  size  of  the  conductors  cutting  lines  of 
force  affects  the  pressure  across  the  terminals  of  a  generator. 
The  number  of  lines  cut  per  second  determines  the  voltage 
of  a  generator,  and  the  materials  of  the  plates  and  electrolyte 
determine  the  voltage  of  a  battery  cell. 

When  the  liquid  is  dilute  sulphuric  acid,  one  plate  is 
zinc  and  the  other  copper,  the  electric  pressure  set  up  is 
always  about  one  volt.  The  chemical  action  takes  place 
between  the  zinc  and  the  acid.  When  one  plate  is  zinc, 
the  other  carbon,  and  the  electrolyte  chromic  acid,  the 
E.M.F.  is  always  about  2  volts,  regardless  of  size  of  plates. 
The  chemical  action  in  this  case  takes  place  between  the 
zinc  and  the  chromic  acid. 

The  important  thing  to  remember  is  that,  in  order  to 
have. any  voltage  set  up  between  the  plates  of  a  cell,  the 
plates  must  be  of  different  materials,  and  one  of  them  must 
be  acted  upon  chemically  by  the  electrolyte.  Even  a  dry 
cell  is  not  really  dry.  The  liquid,  however,  is  retained  in 
blotting  paper  or  in  some  filling,  placed  between  the  plates 
and  in  close  contact  with  them.  As  long  as  the  cell  is 
delivering  current,  one  of  the  plates  (the  negative)  is  being 
consumed,  and  the  character  of  the  electrolyte  is  changing. 
If  this  plate  is  leplaced  by  another  of  the  same  kind  when 
it  is  used  up,  and  the  electrolyte  renewed  by  replacing 
it  with  fresh  liquid,  the  cell  is  then  called  a  PRIMARY  CELL. 

If  an  electric  current  from  an  outside  source  be  sent 
through  the  cell  in  the  opposite  direction,  it  will  often 
cause  the  electrolyte  to  deposit  the  consumed  material 


326 


ELEMENTS  OF  ELECTRICITY 


H2S04 


back  on  the  plate.  When  this  method  of  renewing  the 
materials  is  used,  the  cell  is  said  to  be  a  SECONDARY  or 
STORAGE  CELL. 

186.  Action  of  a  Primary  Cell.  Since  the  action  of 
both  Primary  and  Secondary  cells  is  the  same  when  deliver- 
ing current,  it  is  well  to  study  the  primary  cell,  it  being 
the  simpler,  a  little  more  in  detail. 

Perhaps  the  very  simplest  cell  is  one  composed  of  zinc, 
copper  and  sulphuric  acid  diluted  with  water.     Fig.   233 
represents  such  a  cell.      Cu  is  the 
chemical    symbol    for    copper    and 
Zn  for  zinc. 

Water  is  composed  of  two  parts 
hydrogen  and  one  part  oxygen,  and 
accordingly  is  represented  chemi- 
cally by  the  symbol  H2O. 

Sulphuric  acid  is  composed  of  two 
parts  hydrogen,  one  part  sulphur 
and  four  parts  oxygen,  and  is  repre- 
sented by  the  symbol  H2SO4. 

When  the  acid  H2SO4  acts  chemi- 
cally upon  the  zinc,  Zn,  it  is  broken 
up  into  two  parts,  H2  and  SO4. 
Wherever  there  is  chemical  action  there  is  also  electrical  activity. 
In  this  case,  the  two  parts  of  the  acid  are  charged  with 
opposite  kinds  of  electricity  on  being  separated.  The 
SO4  part  is  negatively  charged  and  clings  to  the  zinc  plate, 
giving  up  its  negative  charge  to  that  plate,  and  taking 
to  itself  some  of  the  zinc  plate,  Zn.  This  small  amount 
of  zinc  now  takes  the  place  of  the  H2  and  makes  a  new 
compound,  ZnSO4,  zinc  sulphate,  which  is  dissolved  by  the 
water  in  the  cell,  just  as  ordinary  salt  is  dissolved  in  water. 
Note  that  during  this  process  the  zinc  plate  becomes  nega- 
tively charged. 

The  other  part,  H2,  of  the  acid  is  charged  positively  and 
instead  of  clinging  to  the  zinc  plate,  it  goes  over  to  the  copper 


FIG.  233. — Simple  primary 
battery  cell. 


ELECTROCHEMISTRY 


327 


ZrK, 


plate,  where  it  gives  up  its  positive  charge,  and  being  a 
gas,  rises  in  bubbles  to  the  surface  of  the  acid.  The  copper 
plate  thus  becomes  positively  charged. 

Fig.  234  represents  this  action  of  the  acid  breaking  up 
into  two  oppositely  charged  parts  and  charging  the  plates. 
Note  that  the  positive  charged  part  of  the  electrolyte  goes 
with  the  current  through  the  cell,  while  the  negative  part 
goes  against  the  current.  The  fact  that  the  charges  on  the 
plates  are  of  opposite  kinds  of  electricity  causes  a  differ- 
ence of  potential  of  approximately 
1  volt  between  the  plates.  There- 
fore if  the  copper  plate  is  joined 
to  the  zinc  by  a  wire,  a  current 
will  flow  through  the  wire  from 
copper  to  zinc,  and  the  chemical 
action  will  continue  to  go  on. 
Thus  the  energy  of  the  chemical 
action  is  transformed  into  the 
electrical  energy  of  an  electric 
current.  If  the  plates  are  not 
joined  by  an  electrical  conductor, 
the  chemical  action  goes  on  until  the 
plates  are  charged  and  then  stops. 

The  action  of  all  cells  is  similar  to  this.  The  electrplyte 
is  always  broken  up  into  two  oppositely  charged  parts. 
One  part  gives  up  its  charge  to  one  plate,  the  other  part 
gives  up  its  charge  to  the  other  plate.  The  difference 
of  potential  thus  set  up  depends  entirely  upon  the  com- 
position of  the  plates  and  electrolyte. 

In  a  well-designed  cell,  the  chemical  action  takes  place 
only  when  the  cell  is  delivering  energy.  The  rate  at  which 
electrical  energy  is  delivered  by  the  cell,  then,  determines 
the  rate  at  which  the  zinc  is  consumed;  just  as  the  rate  at 
which  steam  energy  is  delivered  by  a  boiler  determines 
the  rate  of  the  coal  consumption.  Zinc  may,  therefore, 
be  said  to  be  a  fuel,  the  chemical  energy  of  which  is  turned 


FIG.  234. — Action  within  a 
simple  cell. 


328  ELEMENTS  OF  ELECTRICITY 

into  electrical  energy  by  means  of  a  battery  cell,  just  as 
coal  is  a  fuel,  the  chemical  energy  of  which  is  converted 
into  heat  by  means  of  oxygen,  a  furnace,  water,  and  a  boiler. 

If  energy  could  be  obtained  as  cheaply  by  the  consump- 
tion of  zinc  as  by  the  consumption  of  coal,  primary  (-(41s 
and  electric  motors  would  be  in  as  great  evidence  as  boilers 
and  steam  engines. 

187.  Polarization.  Depolarizers.  When  the  hydrogen 
H2  gives  up  its  positive  charge  to  the  positive  plate,  it  is 
likely  to  cling  to  the  plate  and  form  a  layer  of  hydrogen 
gas  around  it.  This  gathering  of  hydrogen  gas  on  the 
positive  plate  is  called  POLARIZATION.  As  hydrogen  gas 
is  a  very  poor  conductor  of  electricity,  this  makes  the 
internal  resistance  of  the  cell  high  and  a  larger  part  of  the 
pressure  set  up  by  the  chemical  action  must  be  used  to  force 
the  current  through  the  cell.  This  leaves  a  smaller  amount 
of  pressure  to  force  the  current  through  any  outside  circuit. 
The  E.M.F.  is  also  reduced,  because  there  is  an  E.M.F. 
set  up  between  hydrogen  and  copper,  which  opposes  the 
E.M.F.  set  up  between  the  zinc  and  sulphuric  acid. 

Thus,  for  these  two  reasons,  the  terminal  voltage  falls 
as  the  cell  becomes  POLARIZED.  Accordingly,  some  method 
must  be  devised  to  clear  the  positive  plate  of  these  hydrogen 
bubbles.  The  various  means  employed  to  thus  DEPOLARIZE 
a  cell  give  rise  to  various  forms  of  cells.  The  most  common 
method  is  to  introduce  into  the  electrolyte  some  chemical, 
called  a  depolarizer,  which  has  a  surplus  of  oxygen  in  it. 
This  oxygen  then  very  easily  unites  with  the  hydrogen 
and  forms  water,  H2O;  two  parts  hydrogen  to  one  part 
oxygen. 

When  the  depolarizer  acts  quickly  and  gets  rid  of  the 
hydrogen  bubbles  as  fast  as  they  are  formed  on  the  plate, 
the  cell  can  be  run  continuously  without  polarizing,  and 
accordingly  without  much  decrease  in  the  terminal  voltage. 
Such  a  jell  is  called  a  CLOSED  CIRCUIT  CELL. 

When  the  depolarizer  acts  more  slowly  and  the  hydrogen 


ELECTROCHEMISTR  Y 


329 


gradually  gathers  on  the  plate  in  spite  of  the  depolarizer, 
the  cell  can  be  used  only  intermittently,  in  order  to  allow 
the  polarizer  time  in  which  to  act  on  the  hydrogen  and 
clear  up  the  positive  plate.  Such  a  cell  is  called  an  OPEN 
CIRCUIT  CELL.  If  an  open  circuit  cell  is  run  continuously, 
the  terminal  voltage  falls  as  explained  above.  This  explains 
the  action  of  a  dry  cell  on  closed  circuits.  See  Fig.  236 
for  the  results  of  such  a  test  on  an  open-circuit  cell.  If  a 
closed  circuit  cell  is  allowed  to  stand  on  open  circuit,  the 
depolarizer  generally  ruins  the  cell  by  causing  certain 
chemical  changes  in  the  electrolyte.  It  is  essential,  then, 
to  use  the  different  types  of  cells  on  the  circuits  for  which 
they  are  intended.  See  any  hand-book  for  a  discussion 
of  the  various  forms  of  each  type. 

188.  Local  Action.  Besides  polarization,  which  takes 
place  on  the  positive  plate,  there  is  also  the  danger  of  LOCAL 
ACTION,  which  usually  takes 
place  on  the-  negative  plate. 
It  is  generally  caused  by  some 
impurity  in  the  materials.  Sup- 
pose, for  instance,  that  there 
is  a  small  piece  of  carbon  in 
the  zinc  plate.  When  the  zinc 
alone  is  put  into  the  acid, 
the  necessary  requirements  for 
a  battery  are  all  present,  two 
different  conductors  and  a 
liquid.  Now,  referring  to  Fig. 
235,  when  the  acid  is  broken 
up  into  its  two  parts  H2  and 

S04,  the  positively  charged  hydrogen,  instead  of  going 
across  the  cell  to  the  copper  and  charging  that  plate, 
merely  goes  to  the  little  piece  of  carbon,  and  gives  up  its 
charge  to  the  carbon.  Thus  a  small  battery  is  set  in  action 
v/hich  consumes  the  zinc  plate  but  produces  no  terminal 
voltage,  inasmuch  as  the  carbon  impurity,  being  in  contact 


FIG.  235. — Local  action. 


330 


ELEMENTS   OF  ELECTRICITY 


with  the  zinc,  causes  a  short  circuit  current  to  flow  from 
the  carbon  spot  to  the  zinc.  To  avoid  this,  it  is  necessary 
to  have  strictly  pure  materials,  or  at  least  to  amalgamate 
the  zinc  plate  with  mercury  in  order  to  cover  up  the 
impurities. 

189.  Tests  on  Primary  Batteries.  In  Fig.  236  are 
shown  the  results  of  tests  made  by  students  at  Pratt  Insti- 
tute on  an  open  circuit  cell.  The  curves  show  clearly 


40 


Time  in  Minutes 
FIG.  236. — Test  of  primary  battery  (dry  cell). 

the  effect  on  an  open  circuit  cell  when  it  is  kept  running 
for  a  considerable  length  of  time. 

The  external  resistance  was  kept  constant  and  readings 
of  the  terminal  voltage,  current,  and  E.M.F.,  were  taken 
every  few  minutes  for  one  hour. 

The  output  in  watts  was  computed  from  the  product  of 
the  terminal  voltage  times  the  amperes. 

The  internal  resistance  was  computed  by  dividing  the 
difference  between  the  E.M.F.  and  terminal  voltage  by 
the  current.  E.M.F.  —  terminal  voltage  =  volts  required 
to  force  current  through  the  cell. 


ELECTROCHEMISTRY  331 

Notice  the  rapid  decrease  in  the  value  of  the  E.M.F., 
terminal  voltage,  current,  and  output.  Also  note  the  con- 
tinued increase  in  the  internal  resistance.  All  of  these 
effects  are  due  to  polarization,  as  explained  above.  The 
cell  was  then  allowed  to  stand  on  open  circuit  for  15  minutes. 
Note  the  fact  that  the  E.M.F.  gradually  recovers.  The 
depolarizer  is  gradually  clearing  up  the  positive  plate. 

190.  Electrochemical  Equivalent.  Electrolysis.  The 
rate  at  which  the  negative  plate,  generally  made  of  zinc, 
is  consumed,  depends  entirely  upon  how  much  current  is 
being  supplied  by  the  battery.  When  no  current  is  being 
supplied,  all  chemical  action  ceases  if  the  me-tal  is  pure,  and 
thus  no  material  is  consumed.  The  negative  plate  thus 
acts  as  fuel  just  as  coal  acts  as  fuel  in  the  firebox  under  a 
boiler.  The  rate  at  which  coal  is  consumed  depends  upon 
how  much  heat  energy  must  be  converted  into  mechanical 
energy  per  hour.  Similarly  the  rate  at  which  the  negative 
plate  is  consumed  depends  upon  how  much  chemical  energy 
must  be  converted  into  electrical  energy  per  hour.  When  one 
ampere  is  drawn  from  the  cell  it  consumes  a  definite  quantity 
of  zinc  per  hour.  If  two  amperes  are  drawn  from  the  cell, 
twice  as  much  zinc  is  consumed  per  hour.  Nowthis  process 
is  reversible.  If  an  ampere  of  current  were  forced  back 
through  the  cell  in  the  opposite  direction,  this  same  amount 
of  zinc  per  hour  would  be  recovered  from  the  electrolyte 
and  deposited  back  on  the  negative  plate  again.  If  two 
amperes  were  forced  through  in  this  opposite  direction, 
twice  the  amount  of  zinc  would  be  deposited  per  hour 
on  the  negative  plate.  This  reversed  process,  in  which 
the  liquid  is  broken  up  and  the  metal  deposited,  is  called 
ELECTROLYSIS. 

The  definite  quantity  of  a  substance  which  is  consumed 
per  hour  when  one  ampere  is  drawn  from  a  cell,  or  which 
is  deposited  per  hour  by  electrolysis  when  one  ampere  is 
caused  to  flow  back  through  a  cell,  is  called  the  ELECTRO- 
CHEMICAL EQUIVALENT  of  the  substance.  The  following 


332 


ELEMENTS  OF  ELECTRICITY 


table  for  the  common  elements  is  taken  from  the  "  Stand- 
ard "   Handbook. 

ELECTROCHEMICAL  EQUIVALENTS  OF  ELEMENTS. 


Element. 

Symbol 

Grams  per 
Ampere  Hour. 

Ampere   Hour 
per  Gram. 

Aluminum  
CoDDer 

Al 
Cu 

0.3369 
1  186 

2.969 
0  843 

Gold  
Hydrogen 

An 
H 

3.677 
0  03759 

0.2720 
26  60 

Iron 

Fe 

JO.  695 

f  1  439 

Lead  

Pb 

1  1  .  042 

3  858 

I  0  .  959 
0  2592 

Nickel 

Ni 

1  094 

0  914 

Nitrogen          

N 

0  261 

3.828 

Oxygen 

o 

0  2983 

3  352 

Silver  

Ag 

4  025 

0  2485 

Zinc 

Zn 

1  219 

0  820 

By  means  of  this  table  we  can  d3termine  the  amount  of 
zinc,  or  of  any  other  metal,  necessary  to  produce  a  given 
current  for  a  given  length  of  time. 

Example  1.  How  many  ounces  of  zinc  would  be  consumed 
in  8  hours  by  a  battery  delivering  20  amperes?  1  oz.  =  28.4  gms. 

From  table,  1.219  gms.  of  zinc  are  consumed  in  1  hour  by  a 
current  of  1  ampere. 

8X20X1.22  gms.  =  195  gms.,  or  6.97  oz.  consumed  in  8  hours  by 
20  amperes. 

Also  by  means  of  this  table,  it  is  possible  to  find  the  amount 
of  copper  or  other  metal  that  would  be  deposited  in  a  given  time 
by  a  given  current  from  a  salt  solution  of  the  metal. 

Example  2.  How  much  copper  would  be  deposited  from  a 
solution  of  copper  sulphate  in  12  hours  by  a  current  of  2  amperes? 

From  table 

1.19= gms.  deposited  in  1  hour  by  1  ampere. 
2X12X1.19  =  28.8  gms.  deposited  in  12  hours  by  2  amperes. 

Problem  1-12.  How  many  grams  of  silver  are  deposited  in 
8  hours  from  a  solution  of  silver  nitrate  by  a  current  of  4.6  amperes? 


ELECTROCHEMISTRY 


333 


Problem  2-12.  The  zinc  plate  of  a  cell  weighs  4  oz.  How 
long  will  it  last  if  the  battery  is  required  to  deliver  current  of  2 
amperes? 

Problem  3-12.  If  it  is  desired  to  deposit  1.2  Ibs.  of  copper  in 
24  hours,  what  current  must  be  used? 

191.  Electroplating.  In  the  process  called  electro-plat- 
ing, advantage  is  taken  of  the  above  fact  that  if  a  current 
is  passed  through  a  solution  containing  the  salt  of  a  metal, 
the  metal  will  be  deposited  on  the  negative  plate.  The 
positive  plate  of  a  vat  used  for  this  purpose  is  called  the 
ANODE;  the  negative  plate,  the  CATHODE.  The  action  is 
as  follows: 

It  is  desired  to  copper  plate  a  piece  of  iron.  See  Fig.  237. 
The  iron  B  is  placed  in  a  solution  of  copper  sulphate,  CuSO4, 
together  with  a  piece  of  copper  A. 
The  current  is  now  forced  through 
the  cell  from  A  to  B,  and  in  pass- 
ing, it  breaks  up  the  copper  sul- 
phate into  a  copper  part,  Cu,  and 
an  acid  part,  S04.  The  copper 
part,  Cu,  being  positively  charged, 
goes  with  the  current,  to  the  nega- 
tive plate  B  and  is  deposited  on  it, 
thus  giving  it  a  coating  of  copper. 

The  acid  part,  SO4,  being  nega- 
tively charged,  as  explained  in  the 
action  of  a  single  cell,  goes  against 
the  current,  to  the  positive  plate  A 
arid  combines  chemically  with  it, 
forming  more  copper  sulphate, 

CuSO4.     Thus  the  copper  plate  is  gradually  dissolved  and 
conveyed  to  the  iron. 

If  it  is  desired  to  nickel  plate  the  iron,  a  piece  of  nickel 
is  used  for  the  ANODE,  or  positive  plate  A,  instead  of  copper, 
and  nickel  ammonium  sulphate  for  the  electrolyte,  instead 
of  copper  sulphate. 


Cu 


Iron 


FIG.  237. — Action  within  an 
electroplating  vat.  Copper 
plating  iron. 


334  ELEMENTS  OF    ELECTRICITY 

192.  Electrotyping.     The   object    of   electrotyping  is  to 
reproduce    the  printers'  set-up    type,  engravings,    etc.      A 
wax  impression  is  first  taken  of  the  set-up  type.     Since 
wax   is   a    non-conductor   of   electricity,    a    thin    coating 
of  graphite  is  now  given  the  wax  mold.     The  whole  mold 
with  its  coating  of  graphite  is  immersed  in  copper  sulphate 
together  with  a  copper   bar.      An   electric  current  is  now 
sent  through  from  the  copper  to  the  graphite.     This  causes 
copper  to  be    deposited  on   the   graphite   just   as   it    was 
deposited  on  the  iron  in  the  cell  of  the  previous  paragraph. 
The  current  is  allowed  to  run  long  enough  to  deposit  a  plate 
of  sufficient  thickness  to  be  handled  safely.     Then  the  wax 
mold  is  removed  and  the  remaining  copper  plate  is  an  exact 
reproduction  of  the  type. 

193.  Refining    of    Metals.      Since   the   metal,   which   is 
deposited  on  the  negative  plate,  is  remarkably  pure,  this 
method  is  often  used  to  separate  the  impurities  from  a  mass 
of  metal.     The  impure  mass  is  made  the  positive  plate. 
When   an   electric  current  is  passed  through  the  cell,  the 
pure  metal  is  gradually  dissolved  by  the   electrolyte   and 
carried   over   to   the   negative   plate.     At  the  end  of  the 
process,  the  negative  plate  is  found  to  consist  of  very  pure 
metal,  since  practically  all  the  impurities   remain   at  the 
positive  plate. 

For  exact  data  and  methods  of  procedure  in  the  above 
processes,  see  "  Electrochemical  and  Metallurgical  Industry  " 
and  "  Transactions  of  American  Electrochemical  Society." 

Problem  4-12.  How  long  will  it  take  to  refine  100  Ibs.  copper 
if  a  current  of  150  amperes  can  be  used? 

Problem  5-12.  An  iron  casting  is  to  be  copper  plated  and  then 
nickel  plated.  The  current  in  each  case  is  to  be  4  amperes.  How 
long  must  it  remain  in  each  vat  in  order  to  have  12  oz.  of  each 
metal  deposited  on  it? 

Problem  6-12.  A  copper  voltameter,  used  to  calibrate  an  amme- 
ter, has  1.5  grams  of  copper  deposited  on  the  cathode  in  1  hour 
and  5  minutes.  What  value  should  the  ammeter  indicate  during 
the  process? 


ELECTROCHEMISTRY  335 

Problem  7-12.  An  ammeter  being  calibrated  by  a  silver  volta- 
meter, registers  4.38  amperes  for  a  current  which  deposits  39.2 
grams  of  silver  in  2  hours  15  minutes.  What  is  percentage  error 
of  the  ammeter  at  this  reading? 

Problem  8-12.  Two  electroplating  vats  are  arranged  in  series, 
one  for  gold  plating  and  the  other  for  silver  plating.  If  a  current 
flowing  through  the  vats  deposits  1  oz.  of  silver  in  a  given  time, 
how  much  gold  is  deposited  at  the  same  time? 

194.  Definition    of    Ampere.      Voltameter.      Since     one 
ampere  flowing  for  one  hour  through  an  electrolytic  vat 
deposits  a  definite  quantity  of  metal  on  the  negative  plate, 
we  can  determine  very  accurately  the  value  of  an  unknown 
current  by  weighing  the  plate  before  and  after  an  accurately 
timed  run.     On  account  of  this  precise  method  of  deter- 
mining the   current   strength,   the   following   definition   of 
ampere  has  been  established: 

AN  AMPERE  is  THE  RATE  OF  FLOW  OF  A  STEADY  CURRENT 
WHICH  DEPOSITS  4.025  GRAMS  OF  SILVER  PER  HOUR  FROM 
A  SOLUTION  OF  SILVER  NITRATE  DILUTED  TO  A  STANDARD 
DENSITY  WITH  DISTILLED  WATER. 

This  measurement  is  made  with  a  device  called  a  VOLTA- 
METER, which  consists  of  a  glass  jar  containing  the  silver 
plates  and  the  silver  nitrate  solution.  Copper  voltameters 
are  also  used,  consisting  of  two  plates  of  copper  immersed 
in  a  copper  sulphate  solution.  By  means  of  a  voltameter 
it  is  possible  to  accurately  calibrate  galvanometers  and 
ammeters,  although  the  method  has  the  disadvantage  of 
requiring  too  long  a  time,  if  many  parts  of  the  scale  are  to 
be  calibrated. 

195.  Electrolytic  Destruction  of  Metal  Water  Mains,  etc. 
It  is  customary  in  electric  railways  to  use  the  track  as  the 
return    circuit.     The  rails,   not  being   insulated    from   the 
ground,  allow  the  current  to  leak  into  the  ground  and  follow 
any  low-resistance  path  it  can  find,  such  as  a  water  or  SL 
gas   main,    back   to    generator,    which  of    course,    is    also 
grounded. 


336  ELEMENTS  OF   ELECTRICITY 

Fig.  238  is  a  diagram  of  this  action.  Where  the  cur- 
rent enters  the  pipe  at  A  no  harm  is  done.  But  at  the 
point  B,  where  the  current  leaves  the  pipe,  generally  near 
the  generator  station,  there  is  all  the  action  that  takes 
place  at  the  positive  plate  in  an  electroplating  vat.  The 
pipe,  being  in  moist  ground,  is  in  contact  with  water  con- 
taining more  or  less  salt,  which  causes  it  to  act  as  an  electro- 
lyte. Thus,  as  the  electric  current  leaves  the  pipe,  there 
is  a  chemical  action  set  up  between  the  pipe  and  salt  water, 
by  which  the  iron  of  the  pipe  is  eaten  away  in  places,  and 
carried  by  the  electric  current  to  some  outside  substance, 
which  for  the  instant  is  acting  as  the  negative  plate. 


*  -  1 


FIG.  ^238.  —  Electrolytic  destruction  of  water  mains. 

In  a  short  time  heavy  iron  pipes  have  been  eaten  through 
by  this  action.  Among  the  various  precautions  taken 
to  avoid  this  destruction  by  electrolysis  are  the  following: 
(1)  The  heavy  bonding  of  the  rails  to  procure  a  return  of 
low  resistance.  (2)  The  use  of  a  second  trolley  wire  for 
the  return.  (3)  Tapping  a  return  line  on  to  the  water 
mains,  to  conduct  back  to  the  generator  whatever  current 
they  may  be  carrying,  without  allowing  it  to  flow  through 
the  water  and  produce  electrolytic  action. 

Method  No.  2  is  the  most  efficient,  but  also  the  most 
expensive  and  is  very  seldom  used. 

196.  Storage  Batteries.  Efficiency.  In  the  ordinary 
primary  cell,  when  the  negative  plate  is  nearly  consumed, 


ELECTROCHEMISTRY  '  337 

it  is  customary  to  replace  it  by  another.  If,  instead  of 
replacing  the  worn-out  negative  plate,  we  send  a  current 
from  an  outside  source  through  the  cell  in  the  reverse 
direction,  and  deposit*  the  metal  back  on  the  negative 
plate,  the  cell  is  then  called  a  STORAGE  BATTERY.  When 
such  a  cell  is  delivering  current  it  is  said  to  be  discharging; 
when  receiving  current,  it  is  said  to  be  charging.  A  storage 
cell  differs  fundamentally  in  no  respect  from  a  primary 
cell.  Any  primary  cell  could  be  used  as  a  storage  cell 
and  have  its  negative  plate  restored  by  electrolysis  as  an 
ordinary  storage  battery  does.  It  is  not  commercially 
economical,  however,  to  do  this  in  the  case  of  any  but  a  few 
cells,  especially  constructed  for  this  purpose. 

Remember  that  a  storage  cell  does  NOT  store  electricity. 
It  stores  nothing  but  chemical  energy.  In  charging,  elec- 
trical energy  is  transformed  into  chemical  energy  and 
stored  in  the  cell;  in  discharging,  this  chemical  energy  is 
changed  back  again  into  electrical  energy. 

Of  course  there  are  losses  in  both  transformations  which 
prevent  an  efficiency  of  100  per  cent.  For  one  thing,  the 
terminal  voltage  applied  to  the  cell  on  charge  must  be  higher 
than  that  supplied  by  the  cell  on  discharge  at  same  current 
rate,  partly  because,  in  each  case,  the  internal  resistance 
of  the  cell  must  be  overcome. 

Charging  voltage  =E.M.F.   of   cell  +IR  drop   of  cell. 

Discharging  voltage  =E.M.F. -of  cell  —  IR  drop  of  cell. 

In  charging  a  cell,  the  current  must  be  sent  in  against 
the  E.M.F.  and  the  resistance  of  the  cell,  while  on  discharge, 
part  of  the  E.M.F.  of  the  cell  must  be  used  to  overcome 
this  internal  resistance  in  forcing  a  current  to  flow  through 
it  in  the  other  direction;  so  that  only  part  of  the  E.M.F.  is 
available  as  terminal  voltage.  In  this  respect  it  is  merely 
like  any  other  electrical  machine.  This  difference  between 
the  terminal  voltage  on  charge  and  discharge  is  further 
increased  by  the  fact  that  the  E.M.F.  itself  is  less  on  discharge 
than  on  charge,  due  to  the  electrolyte  changing  in  density. 


338  ELEMENTS  OF  ELECTRICITY 

Accordingly,  good  storage  cells  average  only  about  75 
per  cent  efficiency  at  normal  load. 

197.  Plants  and  Faure  Types.  Construction.  The  most 
common  type  of  storage  cell  in  use  at  the  present  time  is  the 
lead  cell.  In  any  cell,  there  must  be  two  plates  of  different 
substances  in  an  electrolyte.  In  the  lead  cell,  the  negative 
plate  is  pure  spongy  lead,  Pb;  the  positive  plate  is  lead 
peroxide,  PbO2;  one  part  lead  to  two  parts  oxygen. 

The  electrolyte  is  sulphuric  acid,  H2SO4,  diluted  with 
water,  H20. 

The  lead  peroxide  and  spongy  lead  are  poor  conductors 
of  electricity,  and  are  not  sufficiently  hard  to  be  made  into 
plates;  so  it  is  necessary  to  attach  them  to  frames  of  some 
harder  material  which  is  a  good  conductor  of  electricity. 
The  material  of  this  framework  must  also  have  the  property 
of  not  acting  as  a  third  plate,  otherwise  the  acid  will  produce 
LOCAL  ACTION  between  the  framework  and  the  spongy  lead 
or  between  the  framework  and  lead  peroxide,  wherever  it 
touches  them.  The  material  usually  chosen  for  this  frame- 
work is  an  alloy  of  lead  and  antimony,  which  is  mechani- 
cally strong  and  produces  no  local  action  in  the  presence 
of  the  other  substances  in  the  cell.  The  lead  peroxide  and 
the  spongy  lead  are  called  the  ACTIVE  MATERIALS,  to 
distinguish  them  from  the  framework,  and  they  form  the 
real  plates  of  the  cell.  The  combination  is  called  a  GRID, 
either  positive  or  negative. 

There  are  two  methods  of  attaching  the  active  materials 
to  the  framework,  which  give  rise  to  two  general  types 
of  lead  cells:  1st,  the  Plante*;  2d,  the  Faure. 

In  the  PLANTE  TYPE,  Fig.  239,  the  surface  of  the  frame- 
work itself  is  roughened  and  then  changed  into  the  proper 
active  material  by  a  chemical  process;  that  is,  the  surface 
of  the  negative  grid  is  made  into  spongy  lead;  the  surface 
of  the  positive  grid,  into  lead  peroxide. 

In  the  FAURE  TYPE,  holes  are  punched  in  the  frame- 
work. Lead  peroxide  is  then  forced  into  the  holes  of  the 


ELECTROCHEMISTR  Y 


339 


positive  grid,  and  spongy  lead  into  the  holes  of  the  negative. 
The  Gould  cell  is  an  example  of  the  Plante  type,  and 
the  "  Chloride  Accumulator  "  of  the  Faure  type.  The  Plante 
cell  is  heavier  for  the  same  capacity,  but  it  is  maintained 
that  this  method  of  attaching  the  active  material  to  the 
framework  holds  it  there  more  securely  than  the  other. 


FIG.  239. — Plate  of  storage  battery.     Plante  type. 

198.  Chemical  Action  on  Discharge.  Consider  Fig. 
240.  The  sulphuric  acid,  H2SO4,  acts  chemically  upon 
the  lead  plate,  Pb,  and  is  broken  up  into  positively  charged 
H2  and  negatively  charged  SO4.  The  SO4  unites  with  the 
lead  plate,  forming  lead  sulphate,  PbSO4,  and  gives  up 
its  negative  charge  to  it.  The  hydrogen  carries  its  positive 
charge  to  the  lead  peroxide  plate,  where  it  gives  it  up,  and 


340 


ELEMENTS  OF   ELECTRICITY 


unites  with  the  oxygen  of  the  lead  peroxide,  forming  water 
H2O.  The  sulphuric  acid  in  contact  with  the  peroxide 
plate  is  also  broken  up  into  H2  and  S04.  The  H2  of  this 
portion  of  the  acid  also  unites  with  the  oxygen  of  the  lead 
peroxide  and  forms  more  water.  The  SO4  part  of  the  acid 
instead  of  going  over  to-  the  negative  plate  unites  with  the 
lead,  Pb,  of  the  lead  peroxide  plate  and  forms  lead 
sulphate.  PbSO4,  on  the  positive  plate.  Thus  both  plates 


+     On  Disbarge 


Pb 


H2SOt 


FIG.  240. — Chemical  action  of  lead 
storage  cell  on  discharge. 


FIG.  241. — Chemical  action  of  lead 
storage  cell  on  charge. 


are  being  reduced  to  lead  sulphate  PbSO4.  The  cell  con- 
tinues to  deliver  current  until  the  plates  are  entirely  reduced 
to  lead  sulphate,  when  of  course,  all  action  will  cease,  since 
there  would  be  but  one  kind  of  material  present,  and  a 
battery  requires  two  kinds.  The  practical  limit  of  discharge, 
however,  is  reached  long  before  both  plates  are  completely 
reduced  to  the  same  material. 

Note  two  things  which  are  taking  place  when  a  cell  is 
discharging : 

1st.  The  acid  is  continually  growing  weaker.  This 
results  in  a  lower  E.M.F. 

2d.  The  active  materials,  lead  and  lead  peroxide,  are 
being  replaced  by  lead  sulphate,  which  has  a  much  higher 


ELECTROCHEMISTRY  341 

resistance  and  is  more  bulky  than  the  active  materials. 
It  therefore  takes  up  more  space  in  the  holes  of  the  grids 
and  tends  to  buckle  them;  especially  if  the  cell  is  dis- 
charged very  rapidly  and  the  le-ad  sulphate  forms 
quickly. 

199.  Chemical  Action  in  Charging.  Refer  to  Fig.  241. 
Assume  both  plates  to  consist  of  lead  sulphate,  PbSO4. 
When  a  current  from  an  outside  source  G  is  sent  through 
the  cell,  it  breaks  up  the  wa.ter  which  has  been  formed  during 
DISCHARGE  into  positively  charged  hydrogen  H2  and 
negatively  charged  oxygen  O.  Part  of  the  positively 
charged  hydrogen  is  now  attracted  to  the  negative  plate 
and  unites  with  the  SO4  of  the  lead  sulphate,  forming 
sulphuric  acid,  H2  SO4,  and  leaving  pure  spongy  lead  at  the 
negative  plate.  The  negatively  charged  oxygen  O  flowing 
against  the  current  is  attracted  to  the  positive  plate. 
Here  it  unites  with  the  lead  Pb  of  the  lead  sulphate  PbSO4 
plate  and  forms  lead  peroxide,  PbO2.  The  SO4  part  of  the 
positive  plate  is  finally  united  to  the  rest  of  the  hydrogen 
liberated  when  the  electric  current  broke  up  the  water 
H2O  into  H2  and  O.  This  action  forms  still  more  -sulphuric 
acid,  and  a  positive  plate  of  lead  peroxide.  When  all  the 
lead  sulphate  has  been  changed  over  to  lead  peroxide  and 
pure  lead,  the  battery  is  restored  to  the  state  it  was  in  before 
it  was  discharged,  and  is  now  ready  to  furnish  current  again. 

Note  that  the  acid  has  been  growing  denser  during 
charge,  therefore  the  E.M.F.  must  have  been  increasing  as 
the  charging  continued. 

The  chemical  equations  for  the  different  actions  on  dis- 
charge and  charge  may  be  written  as  follows: 

DISCHARGE 
Pb02  +Pb  +2H2S04  =2PbS04  +2H2O 


CHARGE 


342  ELEMENTS  OF  ELECTRICITY 

or 

DISCHARGING. 

Positive  Negative  Positive  Negative 

Plate  Plate  Plate  Plate 

Pb02  < Pb 

+   < + 

H2    < SO4 

+ 
H2SO4 

Produces  Produces 

PbSO4  PbS04  Pb02  Pb 

+  + 

2H2O  2H2SO4 

200.  Care  of  Lead  Storage  Cells.  From  a  study  of  the 
chemical  action  of  a  storage  cell  and  the  physical  results 
of  this  action,  we  can  determine  what  treatment  such  a 
cell  should  receive  in  order  to  give  the  most  efficient  service. 

In  the  first  place,  it  should  be* said  that  storage  batteries 
should  have  the  care  of  a  conscientious  skilled  attendant. 
They  are  costly  in  the  first  place  and  easily  ruined.  With- 
out proper  care  they  deteriorate  at  an  alarming  rate.  The 
main  points  to  be  avoided  are  the  following : 

(1)  Too  rapid  charging  or  discharging. 

(2)  Use  of  impure  electrolyte. 

(3)  Use  of  too  dense  or  too  light  electrolyte. 

(4)  Over-charging  and  over-discharging. 

The  harm  that  may  result  from  disregarding  these 
points  is  evident  on  inspection  of  the  chemical  actions. 

(1)  Too  Rapid  Charging  and  Discharging.  In  con- 
structing a  storage  cell,  manufacturers  usually  allow  a 
capacity  of  from  50  to  60  ampere  hours  per  sq.ft.  of  positive 
plate  area,  counting  both  sides  of  the  plates.  This  capacity 
is  computed  on  an  8-hour  charge  and  discharge  rate, 
which  means  that  the  NORMAL  rate  of  charge  or  discharge 

50  to  60 
should  be  — — =6  or  7  amperes  per  sq.ft.  of  positive 


ELECTROCHEMISTR  Y  343 

plate  area.  The  NORMAL  rate  of  charge  and  discharge 
may  then  be  defined  as  that  rate  which  will  satisfactorily 
charge  or  discharge  the  cell  in  eight  hours,  and  is  about 
6  or  7  amperes  per  sq.ft.  of  positive  plate  area.  In  all 
problems  in  this  book  6.5  amperes  per  sq.ft.  is  to  be  con- 
sidered normal  current. 

In  order  to  have  as  large  plate  area  as  possible,  both 
the  positive  and  negative  plates  usually  consist  of  several 
grids  joined  in  parallel.  To  distinguish  the  positive  from 
the  negative,  it  is  merely  necessary  to  remember  that  the 
positive  plate  has  one  less  grid  than  the  negative.  The 
color  of  the  positive  plate  is  reddish  brown,  of  the  negative 
plate,  a  dark  gray. 

If  charge  or  discharge  Is  carried  on  more  rapidly  than 
this,  the  ampere  hour  capacity  of  the  cell  is  lowered. 
When  a  too  rapid  charge  takes  place,  the  chemical  action 
is  violent,  and  the  active  elements  are  not  properly  depos- 
ited on  the  plates  and  are  likely  to  form  a  deposit  in  the 
bottom  of  the  jar,  which  may  eventually  rise  high  enough 
to  reach  the  plates  and  short-circuit  them,  if  it  is  not 
removed. 

On  too  rapid  discharge,  lead  sulphate  often  forms  in 
such  great  abundance  in  the  crevices  of  the  plates,  as  to 
cause  them  to  buckle,  as  previously  explained. 

On  the  other  hand,  the  lead  sulphate  may  form  a  layer 
over  the  active  materials  and  raise  the  internal  resistance, 
besides  greatly  decreasing  the  actual  amount  of  active 
materials  available. 

Problem  9-12.  A  lead  storage  cell  has  a  positive  plate  area 
of  4.2  sq.ft.  (a)  What  is  its  normal  discharge  rate?  (6)  What 
is  the  ampere-hour  capacity? 

Problem  10-12.  A  storage  battery  is  required  to  deliver  10 
amperes,  (a)  What  should  be  the  area  of  the  positive  plate?  (6) 
How  many  ampere-hours  would  such  a  cell  have? 

Problem  11-12.  What  should  be  the  area  of  the  positive  plates 
in  a  lead  cell  of  120  ampere-hours  capacity? 


344  ELEMENTS  OF  ELECTRICITY 

(2)  Use  of  Impure  Electrolyte.     Nothing  but  the  purest 
sulphuric  acid  and  distilled   water  should  be  used  for  the 
electrolyte. 

Any  metallic  impurity  in  the  acid  or  water  is  deposited 
on  the  plates  as  a  pure  metal  and  there  causes  local  action, 
which  we  have  seen  consumes  the  material  of  the  cell  with- 
out producing  any  available  energy. 

Any  acid  impurity  which  will  dissolve  the  lead  of  the 
framework  is  objectionable  for  very  apparent  reasons. 

Too  much  care,  therefore,  cannot  be  exercised  in  select- 
ing the  materials  which  go  to  make  up  the  electrolyte. 

(3)  Too    Light   or    Too    Heavy   Electrolyte.     The  specific 
gravity  at  which  a    cell    should  work   depends  somewhat 
upon  the  use  to  which  it  is  to  be  put.     If  the  cell  is  to  be 
used  continuously,  it  can  work  at  a  higher  density  than 
if  it  is  to  stand  unused  for  any  length  of  time. 

It  is  of  advantage  to  have  as  high  a  density  as  is  practicable, 
because  of  the  decreased  internal  resistance  and  the  increased 
E.M.F.  which  the  cell  has  when  the  acid  is  of  high  density, 
as  previously  explained.  But  if  the  cell  is  to  stand  inactive, 
the  strong  acid  tends  to  change  the  active  material  into 
lead  sulphate  more  rapidly  than  if  weaker  acid  were  used. 

When  studying  the  chemical  action  of  the  cell  we  saw 
that  the  acid  grew  lighter  as  the  cell  was  discharged,  and 
heavier  as  the  cell  was  recharged.  One  method,  then, 
of  telling  what  condition  a  cell  is  in,  with  regard  to  charge 
or  discharge,  is  to  test  its  specific  gravity  with  a  hydrom- 
eter. When  fully  charged  it  should  have  a  specific  gravity 
of  from  1.20  to  1.24,  according  to  the  class  of  work  it  is  to 
do,  as  explained  above.  On  discharge  the  specific  gravity 
should  never  fall  below  from  1.185  to  1.195. 

For  example,  suppose  a  cell  had  a  density  of  1.22  when 
fully  charged.  If  at  a  certain  time  it  showed  a  density 
of  1.189,  it  would  not  be  safe  to  discharge  it  any  further. 

(4)  Over-charging   and   Over-discharging.     When    all   the 
active  material  has  been  changed  into  lead  peroxide  and 


ELECTROCHEMISTRY  345 

spongy   lead,   there   is  nothing   gained   in   continuing   the 
charging,  and  a  waste  of  electric  power  is  going  on. 

There  are  three  ways,  in  which  one  may  tell  when  a  cell 
is  fully  charged: 

(1)  The   plates   begin  to   liberate   ga's  in   large   quan- 
tities.    The  cell  is  said  to  "  boil." 

(2)  The   specific   gravity  rises  t'o  the   value   which   it 
should  have  when  fully  charged,  between  1.20  and 
1.24  (see  previous  paragraph). 

(3)  The  impressed  voltage  AT  NORMAL  CHARGING  rate 
of  current  rises  to  about  2.5  volts.     This  is  because 
a  fully  charged  cell  has  a  high  "E.M.F.  which  has 
to  be  overcome  in  order  to  send  a  charging  current 
through  the  cell. 

NOTE.  The  voltage  of  a  cell  on  open  circuit  shows  ABSOLUTELY 
NOTHING  about  its  condition  as  to  charge  or  discharge. 

The  E.M.F.  (voltage  on  open  circuit)  between  lead  and  lead 
peroxide  in  sulphuric  acid  of  1  .'2  density  is  2.2  volts  (approximately) . 
Since  the  E.M.F.  of  a  cell  does  not  depend  upon  the  area  of  the 
plates,  as  long  as  there  is  any  lead  left  on  one  plate  and  lead  peroxide 
on  the  other,  the  E.M.F.  will  be  about  2.2  volts.  Therefore 
always  take  the  voltage  of  a  cell  when  it  is  either  charging  or 
discharging  at  its  NORMAL  rate. 

OVER-DISCHARGE  is  the  one  great  source  of  trouble  in  a 
storage  cell. 

IT  produces  an  excess  of  lead  sulphate,  which  causes  the 
plates  to  buckle.  Buckling  may  either  short-circuit  the 
cell,  which  ruins  it;  or  cause  the  active  materials  to  fall 
off,  which  decreases  the  capacity. 

IT  weakens  the  acid  and  forms  lead  sulphate  in  excess, 
thus  raising  the  internal  resistance  so  that  the  terminal 
voltage  becomes  very  low. 

IT  weakens  the  acid  so  that  the  E.M.F.  is  lowered  to  a 
marked  extent. 

There  are  two  ways  in  which  one  may  tell  when  a  cell 
has  been  discharged  as  much  as  is  practicable: 


346  ELEMENTS  OF  ELECTRICITY 

(1)  The    specific    gravity    falls    to    1.185-1.195.     (See 
previous  paragraph) . 

(2)  The   voltage,    ON    DISCHARGE   AT   NORMAL   RATE, 
falls  to  1.8  volts. 

CAUTION.  NEVER  CONTINUE  DISCHARGING  A  LEAD  CELL 
AFTER  THE  VOLTAGE  FALLS  TO  1.8  VOLTS  AT  NORMAL  DIS- 
CHARGE CURRENT  RATE. 

201.  Curves  of  Charge  and  Discharge.  The  changes 
in  the  physical  properties  of  a  lead  storage  cell  are  best 


HYCAP  EXIDE  CELL 
43.2  Amp. Rate 


1.6 


234 
•Time  in  Hours 

FIG.  242. — Charge  and  discharge  curves-  of  Exide  storage  cell. 


brought  out  by  curves  plotted  from  a  series  of  readings. 
Fig.  242  shows  a  set  of  curves .  plotted  from  data  taken 
on  an  "  Exide"  cell. 

The  charge  and  discharge  took  place  at  normal  rate. 
Note  the  rapid  rise  in  voltage  during  the  first  and  the  last 
hour  of  charge.  During  the  intervening  five  or  six  hours 
the  voltage  was  fairly  constant. 

On  the  discharge,  the  voltage  falls  rapidly  during  the 
last  hour,  but  remains  nearly  constant  during  the  remain- 
ing time.  This  makes  the  storage  cell  a  first  rate  source 
of  constant  potential. 


ELECTROCHEMISTR  Y 


347 


Fig.  243  shows  the  variation,  on  discharge,  in  specific 
gravity,  E.M.F.  and  internal  resistance. 
NOTE: 

(1)  That  the  specific  gravity  of  the  cell  increases  on 
charge,  and  decreases  on  discharge. 

(2)  That   the  internal  resistance  increases  on  discharge 
slowly  at  first  but  rapidly  as  cell  is  nearly  discharged. 

(3)  That  the  E.M.F.  of  the  cell  rises  slightly  on  charge 
and  falls  slightly  on  discharge. 


1.28 


1.26 


.0010 


Time  in  Hours  (Discharge) 


FIG.  243.  —  Change  in  specific  gravity,  E.M.F.,  and  internal  resistance*  during 

discharge. 

A  method  by  which  the  true  ohmic  resistance  R  can  be  found 
is  as  follows:  Take  terminal  voltage  Ft  and  current  /  on  dis- 
charge. Suddenly  open  the  switch  while  the  cell  is  discharging 
and  record  the  value  which  the  voltmeter  registers  on  the  swing 
of  the  needle.  (The  voltmeter  will  read  instantly  E,  the  E.M.F. 
of  the  cellr  and  then  gradually  rise  as  depolarization  takes  place.) 
Then, 


where   R  =  true  ohmic  resistance  of  cell; 
E  =  E.M.F.  -of  cell; 
F!=  terminal  voltage  on.  discharge  ; 
7  =  rate  of  discharge. 

Problem  12-12.  What  is  the  internal  resistance  of  a  cell  which 
has  a  terminal  voltage  of  2.00  volts  when  discharging  at  a  12- 
ampere  rate,  and  E.M.F.  of  2.32  volts? 


348  ELEMENTS  OF  ELECTRICITY 

Problem  13-12.  What  voltage  would  have  to  be  impressed  on 
t-ell  of  Problem  12  in  order  to  charge  it  at  same  rate? 

*Problem  14-12.  A  cell  has  an  internal  resistance  of  .004  ohm. 
What  would  be  the  difference  in  the  terminal  voltage  on  charge  and 
discharge  at  20  amperes? 

Problem  15-12.  A  cell  of  3.6  sq.ft.  positive  plate  area  has  an 
E.M.F.  of  2.2  volts,  and  an  internal  resistance  of  .005  ohm.  What 
are  terminal  voltages  at  normal  rate  of  charge  and  discharge? 

202.  Advantages  in  Use  of  Storage  Cells.     In  general,  the 
uses  to  which  a  storage  battery  may  be  put  are  three : 

(1)  To  help  the  generators  carry  the  "  peak  "    of    the 
load.     That  is,  the  battery  is  charged  when  the  load  is 
light.     Then,  when  the  load  is  too  heavy  for  the  generators 
to  carry  alone,  the  battery  is  placed  in  parallel  with  the 
generators  to  take  part  of  the  load.     This  use  allows  the 
installation    of    much    smaller    generators,    engines    and 
boilers. 

(2)  To  take  care  of  a  long-continued  light  load  and  allow 
the  engine  and  generator  to  be  shut  down. 

The  battery  is  here  charged  while  the  generator  is  tak- 
ing care  of  its  minimum  load,  and  only  thrown  in  when 
the  generator  is  shut  down.  The  engine  is  run  steadily 
at  its  most  efficient  load,  but  no  fuel  is  wasted  in  keeping 
it  running  to  supply  a  light  load.  Often  the  battery  takes 
care  of  its  load  with  no  attendant  present. 

This  use  saves  both  in  the  cost  of  fuel  and  labor. 

(3)  As  an  auxiliary  source  of  supply  in  case  of  accident. 
Many  companies  now  supply  power  under  heavy  forfeits 
to  keep  the  line  "  alive  "  at  all  times.     This  necessitates 
some  reserve  source  of  supply  which  can  be  drawn  upon 
at  a  moment's  notice.     Storage  batteries  meet  this  demand 
wherever  a  source  of  Direct  Current  is  required. 

203.  Use    on    Constant    Potential    Line.     The  voltage 
of  a  storage  cell  is  seen  to  fall  off  as  the  cell  discharges. 
If  a  battery  of  cells  is  to  be  used  to  supply  power  to  lamps, 


ELECTROCHEMISTRY  349 

some  method  must  be  devised  to  keep  the  voltage  across 
the  line  constant. 

204.  Rheostat  Control.  One  method  shown  in  Fig. 
244  is  to  use  a  resistance  R  in  a  series  with  the  cells. 
This  resistance  can  be  cut  out  as  the  voltage  falls. 

Suppose  115  volts  were  required  across  the  lamps  L. 
When  the  cells  were  nearly  discharged,  each  would  have  a 
voltage  of  1.8  volts  on  normal  rate  of  discharge.  There 

would  be  required  then  to  furnish  this  pressure  — —  =64  cells. 

l.o 

When  the  cells,  were  fully  charged  each  would  have  a 
terminal  voltage  of  about  2  volts.  If  64  cells  were  used 


Storage  Cells 


U| SWWW^- 


FIG.  244.— Rheostat  control  of  cells. 

the  line  voltage  would  then  be  64X2=128  volts,  which 
is  128—115  =  13  volts  in  excess  of  pressure  required.  A 
resistance  R  is  therefore  generally  inserted  between  the 
line  and  the  fully  charged  cells  which  will  take  up  this  13 
volts  in  the  IR  drop  across  it  and  leave  only  115  volts 
across  the  line. 

As  the  voltage  gradually  falls  on  discharge,  an  operator 
from  time  to  time  cuts  out  some  of  this  resistance  R  and 
keeps  the  voltage  across  the  line  constant. 

205.  End  Cell  Control.  Another  method  is  to  cut  in 
and  out  a  certain  number  of  the  cells  at  the  end  of  the 
series,  as  more  or  less  voltage  is  required  by  the  load. 

In  Fig.  245  cells  A,  B,  C,  D,  E,  F,  G  are  called  the  END 
CELLS.  As  the  cells  become  partly  discharged  and  the 
voltage  of  each  falls,  the  arm  M,  of  the  end  cell  switch  is 
moved  out  toward  G,  throwing  more  cells  across  the  line 


350 


ELEMENTS  OF  ELECTRICITY 


and  thus  raising  the  voltage.     (See  any  Electrical  Hand- 
book for  description  of  this  switch.) 

Problem  16-12.  Assume  each  lamp  in  Fig.  244,  takes  10  amp- 
eres at  115  volts.  Assume  terminal  voltage  of  each  of  the  64  cells 
used  in  series  to  be  1.8  volts  when  low  and  2.0  volts  when  fully 
charged.  What  value  would  resistance  R  have  to  be  in  order  to 
keep  the  voltage  across  the  lamps  constant  at  115  volts? 

Problem  17-12.  How  many  end  cells  might  be  used  in 
Problem  16  instead  of  resistance  R? 

Problem  18-12.  Assume  each  cell  to  have  an  internal  resistance 
of  .002  ohm,  and  E.M.F.  of  each  cell  to  range  from  2.2  to  2.0 
volts.  How  many  cells  would  have  to  be  used  to  maintain  110 


FIG.  245. — End  cell  control. 

volts  across  a  set  of  lamps  as  in  Fig.  244,  if  each  lamp  takes  3 
amperes? 

Problem  19-12.  If  a  resistance  R  were  used  to  control  voltage 
in  Problem  18,  what  value  would  it  have  to  be? 

Problem  20-12.  How  many  end  cells  could  be  used  in  Prob- 
lem 18  to  maintain  constant  voltage  at  lamps? 

206.  Methods  of  Changing  Storage  Cells.  In  charging 
a  battery  where  the  END  CELL  method  of  control  is  used, 
all  the  cells  are  thrown  in  at  first.  Since  the  end  cells  are 
used  a  very  much  shorter  time  than  the  rest,  they  are  not 
discharged  to  any  great  extent,  and  require  very  little 
charging.  As  soon  as  one  of  the  end  cells  becomes  fully 


ELECTROCHEMISTR  Y  35  1 

charged  (as  determined  by  methods  previously  described) 
it  is  cut  out  by  moving  the  switch  arm  back  toward  A, 
Fig.  245. 

It  has  been  shown  that  the  voltage  required  to  charge  a  cell 
is  always  greater  than  the  terminal  voltage  of  the  cell  on  dis- 
charge, mainly  on  account  of  the  internal  resistance  of  the  cell. 

The  equations  previously  given,  showing  the  voltage  which  a 
cell  will  give  on  discharge  and  the  voltage  which  is  required  to 
charge  it,  bring  out  this  fact  plainly. 

(1)  VC=E+IRI; 

(2)  Vd  =  E-IRl) 

where  Vc=  terminal  voltage  required  to  charge; 
Vd  =  terminal  voltage  on  discharge  ; 
#  =  E.M.F.  of  cell; 
/=  current; 
RI  =  internal  resistance  (virtual). 

If  we  subtract  (2)  from  (1), 


That  is,  the  voltage  required  to  charge  is  greater  than  the 
terminal  voltage  on  discharge  by  twice  the  IRi  drop  of  the  cell. 

Now  since  the  charging  voltage  is  always  higher  than  the 
discharge  voltage,  it  has  been  shown  that  some  pressure 
higher  than  the  line  pressure  must  be  used  to  charge. 
If  the  line  can  be  disconnected  from  the  generator  dur- 
ing charge,  then  the  voltage  of  the  generator  can  be  raised 
sufficiently  to  get  the  required  charging  pressure.  But 
if  the  generator  must  supply  the  line  and  charge  the 
battery  at  the  same  time,  a  small  low-voltage  generator, 
designed  to  carry  heavy  current,  is  often  connected  in 
series  with  the  main  generator  and  batteries.  This  small 
generator  is  called  a  BOOSTER.  Its  purpose  is  to  raise  the 
voltage  across  the  cells.  Fig.  246  shows  the  method  of 
using  a  booster. 

Assume  the  generator  G  must  maintain  115  volts  across 
the  line  and  at  the  same  time  charge  battery  S,  of  64  cells. 


352  ELEMENTS  OF   ELECTRICITY 

The  battery  would  require  from  128  to  160  volts  to  charge 
(depending  largely  upon  the  degree  of  discharge).  The 
small  generator  B  is  put  in  series  with  the  generator  across 
the  battery  and  raises  the  voltage  enough  to  charge  the 
cells. 

There  are  many  types  of  automatic  and  hand  control 
b  osters,  for  which  see  any  book  on  Electrical  Engineering. 

For  small  installations,  the  battery  is  usually  divided 
into  two  or  three  parallel  parts  during  the  charging.  The 
excess  voltage  of  generator  can  then  be  taken  up  by  a  rhe- 
ostat. On  discharge,  of  course,  the  parts  are  again  connected 
in  series. 


FIG.  246.— Booster  control. 

207.  Floating  Batteries.  There  are  certain  conditions 
under  which  storage  batteries  may  be  used  continuously 
on  the  line  and  be  charged  and  discharged  automatically 
without  any  special  apparatus  such  as  a  booster,  etc. 

Suppose  the  batteries  are  placed  directly  across  the  line 
at  some  distance  from  the  generator,  or  in  such  a  way  that 
there  is  considerable  resistance  in  the  line  wires  between 
the  generator  and  batteries.  This  resistance  would  cause 
a  large  line  drop  when  a  heavy  current  was  sent  through  the 
line.  If  the  line  drop  were  enough  to  cause  the  impressed 
voltage  across  the  cells  to  fall  below  the  E.M.F.  of  the  cells, 
they  would  begin  to  discharge  current  into  the  line  and  so 
aid  the  generator.  If  only  a  small  current  is  sent  through 
the  line,  the  line  drop  would  be  small  and  the  impressed 


ELECTROCHEMISTRY  353 

voltage  across  the  cells  would  be  higher  than  their  E.M.F. 
The  cells  would  then  charge  and  take  current  from  the 
line. 

In  Fig.  247  there  is  .80  ohm  in  the  line  (trollej^  and  track) 
from  generator  to  car  No.  1.  When  the  car  is  taking  100 
amperes,  there  is  a  line  drop  of  .80X100=80  volts.  The 
voltage  across  the  line  at  this  point  is  then  550  —  80=470 
volts. 

Assuming  the  E.M.F.  of  one  cell  to  be  about  2.2  volts, 
if  we  place  a  battery  of  220  cells  across  the  line  at  this 
point,  we  should  have  an  E.M.F.  of  220X2.2=484  volts; 
that  is,  14  volts  (484  —  470)  more  than  the  impressed  voltage. 
The  battery  would  then  discharge  into  the  line  and  help 
the  generator  to  furnish  current  to  the  three  cars. 


B 

FIG.  247. — Floating  battery  on  railway  line. 

If,  however,  car  No.  1  were  taking  only  10  amperes,  and 
cars  Nos.  2  and  3  were  taking  no  current,  there  would  be  a 
line  drop  of  only  10 X. 80  =8  volts.  The  voltage  across  the 
points  AB  would  then  be  550  —  8=542  volts,  which  is  58 
volts  (542-484)  higher  than  t'he  E.M.F.  of  the  storage 
battery  across  the  line  at  that  place.  The  cells  would  then 
charge  and  draw  current  from  the  line.  (Of  course  in 
taking  more  current  from  the  line  the  cells  would  cause 
a  still  greater  drop  in  the  line  voltage  from  generator  to 
cells,  so  that  there  would  exist  considerably  less  than  the 
58  volts  difference  between  the  battery  E.M.F.  and  the 
line  voltage.) 

Batteries  can  only  be  "  floated  "  at  a  point  where  there 
is  a  large  fluctation  in  the  voltage  impressed  across  them, 
such  as  railway  lines,  etc. 


354 


ELEMENTS  OF  ELECTRICITY 


Problem  21-12.  In  a  system  arranged  as  in  Fig.  247,  the  gen- 
erator brush  potential  is  565  volts.  Trolley  is  No.  00  copper 
wire.  Track  has  resistance  of  .03  ohm  per  mile.  How  far  from 
generator  should  a  set  of  224  lead  storage  cells  be  placed  in  order 
to  "  float  "  satisfactorily?  Maximum  current  in  section  out  to 
cells  is  150  amperes.  Average  E.M.F.  of  each  cell  2.1  volts. 

Problem  22-12.  In  a  system  arranged  as  in  Fig.  248:  gen- 
erator has  a  resistance  of  .2  ohm,  E.M.F.  of  120  volts.  R  =  A 
ohm.  B  consists  of  56  cells  of  2  volts  E.M.F.  How  many  amperes 
must  load  use  in  order  that  the  battery  B  may  deliver  current 
to  the  line?  Neglect  line  resistance. 


FIG.  248. 


208.  Edison  Storage  Battery.  A  storage  battery  invented 
by  Thomas  Edison  and  manufactured  by  the  Edison  Storage 
Battery  Company  has  recently  appeared  on  the  market. 
It  is  designed  especially  for  traction  work,  being  light  and 
compact. 

The  positive  plate  consists  at  first  of  nickel  hydrate, 
NiOH2,  held  in  steel  tubes  firmly  attached  to  a  steel  grid. 

The  negative  plate  consists  of  iron  oxide,  FeO,  held  in 
pockets  on  a  metal  plate. 

The  electrolyte  is  a  21  per  cent  solution  of  caustic  potash, 
KOH,  in  pure  distilled  water.  Specific  gravity  1.20. 

The  retaining  jar,  instead  of  being  either  glass  or  vulcanite 
as  in  the  case  of  other  storage  cells,  is  made  of  nickel-plated 
sheet  steel.  The  chemical  actions  which  take  place  within 
the  cell  are  not  yet  clearly  understood.  The  following  are 
supposed  to  approximate  the  reactions; 


ELECTROCHEMISTRY 


355 


When  the  cells  are  first  charged,  the  nickel  hydrate, 
NiOH2,  is  changed  to  nickel  oxide  NiO2,  and  the  iron 
oxide,  FeO,  is  reduced  to  metallic  iron,  Fe.  The  cell  may 
thus  be  considered  to  consist  of  a  positive  plate  of  nickel 
oxide,  NiO2,  and  a  negative  plate  of  pure  iron,  Fe. 

209.  On  Discharge.  The  nickel  oxide,  NiO2;  probably 
goes  to  a  lower  oxide  of  nickel,  Ni2O3;  that  is,  there  are 
only  three  parts  oxygen  to  two  of  nickel,  instead  of  two 
oxygen  to  one  of  nickel.  The  oxygen  thus  liberated  is 
negatively  charged  and  goes  over  to  the  iron  plate,  gives 
up  its  negative  charge,  and  unites  with  the  plate  to  form  iron 


FeO 


FIG.  250. — Action  within  Edison  eel 
on  discharge. 


FIG.  251. — Action  within  Edison  cell 
on  charge. 


oxide,  either  FeO  or  Fe2O3.  The  electrolyte  neither  loses 
nor  gains  in  the  process  and  therefore  remains  at  the  same 
density  throughout  discharge  and  charge.  The  potassium 
hydroxide  seems  to  act  merely  as  a  catalyzer  or  carrier. 
Fig.  250  shows  the  action  on  discharge. 

210.  On  Charge.  The  iron  oxide,  FeO,  is  broken  up  and 
the  negatively  charged  oxygen  leaves  the  iron  plate, 
travels  back  against  the  current  through  the  cell  and  unites 
with  the  positive  nickel  oxide,  Ni2Os,  and  forms  a  higher 
oxide,  NiO2.  Fig.  251  shows  this  action  during  charge. 

The  following  equations  show  at  a  glance  the  chemical 
reaction : 


356  ELEMENTS  OF  ELECTRICITY 

DISCHARGE 

Fe  +2NiO2  +  H2O  -f  KOH  =  FeO  +  Ni2O3  +  H2O  +KOH. 

CHARGE. 
DISCHARGE.  CHARGE. 

Positive  Negative  Positive  Negative 

Plate  Plate  Plate  Plate 

Ni02)                   Fe                              Nia°»  Fe° 

-f-                         -f                                 +•  + 

H2      < O  O >  H2 


KOH  KOH 

Produces  Produces 

Ni203  FeO  {*gJ2  Fe 


H20  H20 

KOH  KOH 

211.  Physical  Changes.  As  stated  above,  the  specific 
giavity  of  the  electrolyte  does  not  depend  upon  the  state 
of  charge  or  discharge.  Some  other  method,  then,  must 
be  used  to  determine  the  condition  of  this  form  of  storage 
cell  as  to  charge  and  discharge. 

Fig.  252  shows  some  curves  plotted  from  data  taken 
on  an  Edison  storage  cell,  type  "A6." 

Note  that  on  discharge  the  voltage  falls  very  rapidly 
at  first  as  in  a  lead  cell.  Unlike  a  lead  cell,  however,  there 
is  no  voltage  at  which  it  remains  nearly  constant,  but 
has  a  general  downward  trend.  Near  the  point  of  complete 
discharge  the  voltage  falls  off  rapidly  again,  as  in  a  lead 
cell.  The  condition  may  thus  be  determined  closely  by  the 
voltage  on  discharge. 


ELECTRO  'HEMISTRY 


357 


The  average  terminal  voltage  on  discharge  at  normal 
current  is  1.2  volts.  The  lowest  voltage  is  1.00  volt. 

The  voltage-  curve  on  charge  is  slightly  irregular,  but 
has  the  peculiar  characteristic  of  reaching  a  maximum  of 
1.84  volts  on  normal  current  when  fully  charged.  Average 
voltage  on  charge,  1.7  volts. 

212.  Comparison  of  Lead  and  Edison  Cells.  Average 
voltage  per  cell  on  discharge:  Lead  cell,  2.00  volts;  Edison 
cell,  1.2  volts.  Watt  hours  per  Ib.  of  cell  complete:  Lead, 
8.5  watt-hours;  Edison,  16.8  watt-hours. 

Internal  resistance :  The  Edison  cell  has  a  slightly  higher 
internal  resistance;  the  efficiency  is  therefore  lower. 


345 
Time  in  Hours 

FlS.  252. — Change  in  terminal  voltage  on  charge  and  discharge. 

The  "  life  "  of  the  Edison  cell  under  commercial  condi- 
tions is  as  yet  undetermined. 

It  is  claimed  for  the  Edison  cell  that  it  can  stand  for  an 
indefinitely  long  time,  charged  or  discharged,  and  not  be 
injured.  A  lead  cell  cannot  remain  any  time  at  all  un- 
charged, and  only  a  few  months  charged,  without  being 
ruined. 

The  temperature  range  at  which  the  Edison  cell  will 
work  best  is  limited  to  the  region  around  70°  F.  Any 
condition  of  use  which  tends  to  raise  or  lower  the  tempera- 
ture decreases  the  efficiency  to  a  marked  extent.  The  lead 
cells  are  not  so  limited  in  this  respect.1 

1  For  further  information  concerning  lead  cells,  the  student  is 
referred  to  Lamar  Lyndon's  "  Storage  Battery  Engineering." 


358  ELEMENTS  OF  ELECTRICITY 


SUMMARY  OF  CHAPTER  XII 

ELECTRIC  BATTERIES.  Devices  for  transforming  chem- 
ical energy  into  electrical  energy.  Require  two  unlike  con- 
ductors, called  positive  and  negative  plates,  and  an  electro- 
lyte which  acts  chemically  upon  one  of  the  plates. 

PRIMARY  CELL.  An  electric  battery  in  which  worn- 
out  element  is  replaced  by  another.  Negative  element  gen- 
erally of  zinc. 

STORAGE  CELLS.  An  electric  battery  in  which  worn-out 
element  is  replaced  by  the  electrolytic  action  of  an  electric 
current  forced  through  cell  in  reverse  direction.  Elements 
generally  of  lead  and  lead  compounds. 

ACTION  IN  CELL.  By  the  chemical  action  taking  place  at 
the  negative  plate,  the  electrolyte  is  broken  up  into  two  oppo 
sitely  charged  parts.  The  positively  charged  part  gives  up  its 
charge  to  positive  plate,  and  negatively  charged  part  gives 
up  its  charge  to  negative  plate.  A  difference  in  potential 
between  the  plates  is  thus  produced. 

POLARIZATION.  Hydrogen  bubbles  collect  on  the  posi- 
tive plate  and  increase  the  resistance  of  cell,  and  lower  the 
E.M.F.  by  setting  up  a  counter  E.M.F. 

DEPOLARIZATION.  Some  oxydizing  agent  is  used  as 
a  depolarizer  to  remove  the  hydrogen  by  uniting  with  it  to 
form  water.  Cells  which  are  depolarized  rapidly  are  used 
for  closed  circuit  work,  those  which  are  depolarized  more 
slowly,  for  open  circuit  or  intermittent  work. 

LOCAL  ACTION.  Any  impurities  in  the  negative  plate 
cause  a  difference  of  potential  to  exist  between  plate  and 
impurity,  forming  a  short  circuit  current,  which  consumes 
the  plate,  but  produces  no  available  energy. 

ELECTROLYSIS.  When  an  electric  current  is  sent  through 
a  solution  containing  a  metal  salt,  it  will  deposit  the  metal 
on  the  negative  plate.  If  chemical  action  takes  place  between 
the  electrolyte  and  positive  plate,  the  positive  plate  is  con- 
sumed. 

ELECTROCHEMICAL  EQUIVALENT.  The  amount  of 
metal  deposited  on  the  negative  plate,  and  the  amount  taken 
from  the  positive  plate,  by  an  ampere-hour  of  electricity  is 
a  constant,  depending  upon  metals  and  electrolyte. 

ELECTROPLATING.  Use  is  made  of  electrolysis  to 
plate  conducting  materials;  to  refine  metals;  to  make  elec- 


ELECTROCHEMISTR  Y  359 

trotypes,  etc.,  and  to  restore  the  active  elements  in  a  storage 
battery. 

DESTRUCTION  OF  WATER  MAINS,  ETC.,  BY  ELEC- 
TROLYSIS. The  leakage  currents  from  the  return  circuits 
in  electric  railways  travel  along  iron  pipes,  and  are  likely  to 
eat  away  portions  of  the  iron  at  the  point  where  they  leave 
the  pipe  to  return  to  the  generator. 

STORAGE  CELLS.  Do  not  store  electricity,  but  chemical 
energy.  Common  types  composed  of  lead  peroxide,  positive 
plate;  and  spongy  lead,  negative  plate. 

Action  on  charge  and  discharge: 

Discharge. 


Pb02  +  Pb  +  2H2S04  =  2PbSO4  +  2H2O 
Charge. 


CARE   OF   STORAGE   CELLS.     Storage   cells   are   injured 
by: 

(1)  Too  rapid  charging  or  discharging. 

(2)  Use  of  impure  electrolyte. 

(3)  Use  of  too  dense  or  too  light  electrolyte. 

(4)  Over-charging  and  over-discharging. 
CONDITION  OF  CELLS.     Can  be  ascertained: 

(1)  By  color  of  plates. 

(2)  By  specific  gravity  of  electrolyte. 

(3)  By  terminal  voltage  at  normal  charging  or  discharg- 
ing rate    (not  by  the  E.M.F.). 

USE  OF  STORAGE  CELLS. 

(1)  To  help  carry  "  peak  "  of  load. 

(2)  To  carry  all  the  "  light  load." 

(3)  As  a  reserve  supply  of  electrical  energy. 
PRACTICE  IN  USE.     Voltage  across  cells  may  be  controlled 

by: 

(i)  Resistance  in  series   with   cells, 
(a)  End  cells. 

(3)  Floating  the  cells  on  the  line  at  a  point  of  consider- 
able voltage  fluctuation. 

EDISON     STORAGE     CELLS.     Positive     element,     nickel 
oxide,  changes  to  lower  oxide  on  discharge. 

Negative  element,  spongy  iron,  changes  to  iron  oxide  on 
discharge. 


360  ELEMENTS  OF  ELECTRICITY 

Electrolyte,  21  per  cent  solution  of  potassium  hydrate, 
which  remains  at  same  specific  gravity  throughout  charge 
and  discharge. 

Is  lighter  per  watt-hour  than  most  lead  cells,  but  has  lower 
E.M.F.  per  cell.  Can  remain  charged  or  discharged  indefi- 
nitely without  deteriorating.  Efficient  temperature  range  is 
limited. 


PROBLEMS  ON  CHAPTER  XII 

23-12.  How  long  must  a  current  of  800  amperes  run  in  a  copper 
refining  vat  to  deposit  enough  copper  to  make  1000  ft.  of  No.  0 
trolley  wire? 

24-12.  It  is  desired  to  copper  plate  an  iron  casting  having 
2.4  sq.ft.  of  surface.  How  long  must  a  current  of  40  amperes 
run  in  order  to  make  the  plate  ^  in.  thick? 

25-12.  Zinc-coated  iron  is  called  "galvanized  iron."  An  iron 
sheet  4  ft.  square  is  to  be  "  galvanized."  A  current  of  30  amperes 
running  for  2  hours  will  deposit  how  thick  a  plate  of  zinc  on  the 
iron? 

26-12.  A  primary  cell  using  zinc  as  negative  plate  maintains 
a  terminal  voltage  of  1.0  volt  at  normal  rate  of  discharge.  A 
steam  plant  has  an  efficiency  of  6  per  cent.  Zinc  costs  5  c.  per 
Ib.  Coal  $5.00  per  ton.  What  is  relative  cost  per  K.  W.-hour 
of  zinc  and  coal  as  fuels? 

27-12.  A  battery  of  20  lead  storage  cells  in  series,  each  having 
2  sq.ft.  positive  plate  area,  is  to  be  charged  from  a  110-volt 
generator.  How  much  resistance  must  be  placed  in  series  with 
the  cells,  assuming,  2.2  E.M.F.  and  .005  ohm  resistance  for  each 
cell? 

28-12.  If  the  cells  in  Problem  27  are  to  be  charged  from  a  110- 
volt  circuit  in  two  parallel  sets  of  10  cells  in  series,  what  series 
resistance  is  needed? 

29-12.  A  set  of  80  lead  storage  cells  each  having  2.2  volts 
E.M.F.  and  .003  ohm  internal  resistance,  is  to  be  charged  in  two 
parallel  sets  of  40  cells  in  series.  Each  cell  has  positive  plate  area 
of  4.2  sq.ft.  What  must  be  terminal  voltage  of  generator? 


ELECTROCHEMISTRY  361 

30-12.  If  cells  in  Problem  29  are  discharged  in  series  at  normal 
rate,  what  will  be  terminal  voltage  of  set? 

31-12.  What  must  be  the  E.M.F.  of  generator  in  Problem  29, 
if  its  internal  resistance  is  .02  ohm? 

32-12.  If  generator  in  Problem  27  has  an  E.M.F.  of  112  volts 
and  an  internal  resistance  of  .32  ohm,  what  series  resistance  must 
be  used  in  charging  the  cells? 

33-12.  If  a  generator  of  135  volts  and  .03  ohm  internal  resist- 
ance be  used  in  Problem  29,  what  resistance  must  be  placed  in  series 
with  the  cells? 

34-12.  If  generator  in  Problem  48-8  has  its  field  separately 
excited  from  110  volts  circuit,  at  what  speed  must  it  be  run  in 
order  to  charge  a  set  of  64  storage  cells  arranged  in  two  parallel 
sets  of  32  cells  in  series?  Each  cell  has  an  internal  resistance 
of  .004  ohm  and  an  E.M.F.  of  2.3  volts.  Normal  current  is  25 
amperes  per  cell. 

35-12.  (a)  What  is  the  ampere-hour  capacity  of  set  of  cells 
in  Problem  27?  (6)  What  is  the  K.W.-hour  capacity? 

36-12.  (a)  What  is  the  ampere-hour  capacity  of  each  cell  in 
Problem  29?  (6)  What  is  the  K.W.-hour  capacity  of  set? 

37-12.  A  set  of  90  lead  storage  cells,  each  having  2.2  volts 
E.M.F.  and  .004  ohm  internal  resistance,  is  to  be  charged  from 
a  110-volt  line.  If  each  cell  is  to  take  its  normal  current  of  20 
amperes,  what  would  be  best  arrangement  of  cells  in  order  to  have 
least  power  lost  in  a  series  resistance  in  line? 

38-12.  How  could  cells  in  Problem  37  be  arranged  in  order  to 
deliver  120  amperes  and  not  exceed  normal  current  of  each  cell? 
At  what  voltage  would  they  deliver  this  current? 

39-12.  A  lead  storage  cell  delivers  20  amperes  at  2.1  volts. 
It  requires  2.4  volts  to  charge  this  cell  at  20  ampere  rate.  What 
is  the  virtual  resistance  of  cell? 

40-12.  A  storage  battery  of  240  cells  in  series  is  "  floated  "  at 
the  end  of  a  4-mile  trolley  line,  the  resistance  of  which  (line  arid 
return)  is  .08  ohm  per  mile.  Generator  voltage  is  550  volts; 
each  cell  has  an  E.M.F.  of  2.1  volts  and  an  internal  resistance  of 
.001  ohm.  What  current  will  cells  supply  to  the  line,  when  there 
are  5  cars  at  the  batteiy  end  of  the  line,  each  taking  65  amperes? 

41-12.  What  will  be  terminal  voltage  of  the  set  of  battery  cells 
in  Problem  40? 


362  ELEMENTS  OF  ELECTRICITY 

42-12.  What  current  will  cells  in  Problem  40  take  when  there 
is  no  load  on  the  line? 

43-12.  What  current  will  generator  be  delivering  when  the  5 
cars  of  Problem  40  are  half  way  between  station  and  battery? 

44-12.  What  current  will  battery  be  delivering  to  or  receiving 
from  line  in  Problem  43? 

45-12.  If  there  are  but  2  cars  on  line  in  Problem  43  each  taking 
75  amperes,  (a)  What  power  is  generator  delivering?  (b)  What 
power  is  battery  receiving  or  delivering? 


CHAPTER  XIII 
PHOTOMETRY   AND   ELECTRIC   ILLUMINATION 

Illumination;  Distribution,  Color,  Intensity — Law  of  Inverse  Squares 
— Photometry,  Candle  Power;  Horizontal  and  Spherical — Foot- 
Candle — Use  of  Sharp-Millar  Illuminometer — Arc  Lamp;  Ballast 
and  Regulating  Resistances — Flaming  or  Luminous  Arcs — Incan- 
descent Lamps,  Carbon,  Tantalum,  Tungsten,  Nernst — Life; 
Effect  of  Overburning;  of  Underburning — Mercury  Vapor  Lamp 
— Moore  Tube. 

IN  taking  up  the  subject  of  illumination,  it  is  perhaps 
well  to  start  with  some  general  ideas  underlying  the  whole 
problem.  In  the  first  place,  it  should  be  stated  that  we 
illuminate  objects  for  the  sole  purpose  of  making  them 
visible  to  the  eye.  The  eye,  then,  is  the  natural  starting 
point. 

When  passing  upon  the  merits  of  any  scheme  of  ordinary 
illumination,  that  which  should  mark  it  as  a  success  or 
failure,  should  be  the  general  effect  of  the  scheme  on  the 
eye.  Success  should  be  measured  largely  by  the  degree 
of  clearness  with  which  the  objects  are  perceived  by  the 
eye,  as  to  shape  and  color.  If  certain  parts  of  a  room  or 
street  are  too  brilliantly  lighted,  objects  in  the  dimmer 
portions  are  not  perceived  by  the  eye.  If  a  certain  side 
of  one  object  be  too  highly  illuminated,  the  general  shape 
of  the  object  is  lost,  as  the  eye  does  not  readily  perceive 
its  more  dimly  lighted  parts.  This  is  because  the  eye 
automatically  adjusts  itself  to  the  most  brilliantly  lighted 
area  within  its  view,  and  accordingly,  is  out  of  adjustment 
for  perceiving  the  rest.  We  should  get  rid  of  the  idea, 

363 


364  ELEMENTS  OF  ELECTRICITY 

therefore,  that  a  light  of  intense  brilliancy  is  the  thing  to 
be  sought.     It  is,  in  general,  highly  undesirable. 

The  problem,  then,  resolves  itself  into  two  parts :  The 
first  object  should  be  to  secure  a  kind  of  lamp  which  will 
cause  objects  to  appear  in  their  accustomed  colors;  that 
is,  the  colors  in  which  they  appear  by  sunlight.  The 
second  object  is  to  so  distribute  the  lamps  that  the  several 
illuminated  surfaces  receive  their  share  of  the  light,  and 
yet  no  bright  light  is  thrown  directly  into  the  eyes. 

213.  Nature  of  Light.  All  space  is  supposed  to  be 
filled  with  a  medium  infinitely  lighter  than  air,  called  ether. 

The  sensation  of  light  is  experienced  when  certain 
wave  motions  in  this  ether  are  transmitted  to  the  eye. 
These  wave  motions  are  called  light  waves.  Light  waves 
differ  from  one  another  in  length  and  violence.  The 
difference  in  length  causes  a  difference  in  color.  Thus 
short  waves  may  be  blue  or  violet,  while  longer  waves  may 
be  red  or  orange.  If  we  have  a  source  of  light  which  sends 
out  long  ether  waves,  we  may  expect  a  predominance  of 
red  and  orange  light  in  it.  The  sunlight  contains  waves 
of  practically  all  lengths  and  thus  is  composed  of  all  colors. 
The  difference  in  violence  of  the  waves  gives  rise  to  a  differ 
ence  in  intensity  of  the  light. 

When  these  light  waves  strike  any  object,  they  are  partly 
reflected  and  partly  absorbed.  Substances  differ  widely 
as  to  the  percentage  of  light  they  absorb  and  the  percentage 
they  reflect.  If  two  objects  are  illuminated  by  the  same 
amount  of  light,  the  one  which  absorbs  the  less  light  and 
reflects  the  more  will  appear  the  brighter. 

Some  objects  reflect  light  waves  of  a  certain  length  only, 
and  absorb  all  the  rest.  It  is  this  property  that  gives  color 
to  objects.  Suppose,  for  instance,  that  a  piece  of  cloth 
were  receiving  light  from  the  sun,  all  of  which  it  absorbed 
except  the  waves  of  proper  length  to  cause  a  sensation  of 
green  to  the  eye.  The  green  waves  only  would  then  come 
from  the  cloth  to  the  eye,  all  the  rest  being  absorbed,  and 


PHOTOMETRY  AND  ELECTRIC  ILLUMINATION    365 

the  cloth  would  appear  green.  If  it  absorbed  waves  of 
all  lengths,  it  would  appear  black,  because  no  light  would 
be  reflected  from  it  to  the  eye. 

If  now  the  piece  of  cloth,  which  absorbs  all  wave  lengths 
except  that  of  green,  were  exposed  to  a  source  of  light  which 
was  emitting  all  colors  except  green,  there  being  no  green 
waves  to  be  reflected  from  it,  the  cloth  in  this  light  would 
appear  black. 

Suppose  a  piece  of  cloth  absorbed  all  colors  but  two, 
say  violet  and  red.  When  light  having  all  wave  lengths 
fell  upon  it,  it  would  absorb  all  the  waves  except  violet 
and  red.  These  two,  the  cloth  would  reflect  as  a  mixture 
and  would  appear  purple.  If,  however,  the  source  of  light 
contained  no  violet  waves,  it  could  only  reflect  the  red 
waves  and  appear  red.  This  light,  then,  would  not  cause 
the  cloth  to  show  its  normal  color. 

So  in  choosing  an  artificial  source  of  light,  it  is  necessary 
to  select  one  which  shall  send  out  all  wave  lengths,  if  we 
wish  to  have  the  different  objects  appear  in  their  normal 
colors. 

214.  Light  Intensity.  Candle-power.  Foot-candle.  A 
source  of  light  sends  out  LIGHT  FLUX,  just  as  a  magnetic 
pole  sends  out  MAGNETIC  FLUX.  The  laws  concerning 
INTENSITY  OF  ILLUMINATION  are  similar  to  those  concerning 
INTENSITY  OF  A  MAGNETIC  FIELD. 

Just  as  a  magnetic  pole  is  of  unit  strength  when  it  pro- 
duces a  unit  force  at  a  unit's  distance,  so  a  source  of  light 
is  of  unit  intensity  when  it  produces  unit  intensity  of  illu- 
mination at  a  unit's  distance. 

The  intensity  of  illumination  which  a  standard  candle 
will  throw  upon  a  surface  placed  one  foot's  distance  from 
it  (at  right  angles  to  rays  of  light)  is  the  UNIT  INTENSITY 
of  ILLUMINATION.  This  unit  is  called  a  FOOT-CANDLE. 
The  candle  itself  is  said  to  have  an  intensity  of  one  CANDLE- 
POWER.  A  candle  is  standard  when  its  intensity  is  1.136 
times  the  intensity  of  a  standard  Hefner  lamp  burning 


366  ELEMENTS  OF  ELECTRICITY 

under  standard  conditions.  Thus  a  light  of  16-c.p.  intensity 
would  illuminate  a  surface  placed  one  foot  from  it  with  an 
intensity  of  16  foot-candles.1 

Careful  distinction  should  be  made  between  candle-power 
and  foot-candle.  CANDLE-POWER  is  the  measure  of  the 
intensity  of  a  source  of  light.  The  FOOT-CANDLE  is  the 
measure  of  the  intensity  of  illumination  of  some  surface 
upon  which  the  light  falls. 

Again,  just  as  the  field  intensity  about  a  magnet  varies 
inversely  as  the  square  of  the  distance  from  the  pole,  so 
the  intensity  of  illumination  of  a  surface  varies  inversely 
as  the  square  of  its  distance  from  the  source  of  light. 
This  is  called  the  LAW  OF  INVERSE  SQUARES.  Accordingly, 
a  32-c.p.  lamp  will  illuminate  a  surface  1  ft.  away  from 
it  with  an  intensity  of  32  foot-candles,  but  a  surface  4  feet 

32 
away,  with  only  -^  or  2  foot-candles. 

Rule.  To  find  the  intensity  of  illumination  on  any 
surface  (at  right  angles  to  light  rays)  divide  the  CANDLE- 
POWER  of  the  lamp  by  the  SQUARE  of  the  distance  in  FEET. 
The  result  will  be  FOOT-CANDLE  illumination.2 


1  For  a  description  of  Standard  Hefner  Lamp,  see  Reference  Books 
on  Photometry. 

2  It  is  often  desired  to  find  the  illumination  of  a  surface  which  is 

not  at  right  angles  to  the  rays  of 
light.  Referring  to  Fig.  253:  let 
A  be  a  source  of  light  of  /  candle- 
power,  and  hung  lv  feet  above  the 
floor.  It  is  desired  to  find  the  inten- 
sity of  light  on  the  horizontal  surface 
S.  Let  6  equal  the  angle  which  (I), 
the  ray  of  light  from  A  to  S,  makes 
with  the  vertical  (lv).  Then  E,  inten- 
sity  of  illumination  of  5  in  foot-candles, 
»  can  be  found  from  the  equation 

E  =  — -  cos30.     This  is  a  very  conveni- 
ent equation  for  finding  illumination  of  different  floor  areas,  etc. 


PHOTOMETRY  AND  ELECTRIC  ILLUMINATION    367 


Problem  1-13.  (a)  What  is  the  illumination  on  a  surface  5 
ft.  from  a  32-c.p.  lamp?  (b)  How  far  from  a  16-c.p.  lamp  will 
the  illumination  be  the  same  as  in  (a)? 

Problem  2-13.  The  illumination  required  on  a  printed  page 
for  easy  reading  is  about  2  foot-candles,  (a)  How  high  above 
a  reading  table  should  a  16-c.p.  lamp  be  hung?  (6)  a  32-c.p. 
lamp? 

Problem  3-13.  It  is  desired  to  illuminate  a  space  on  the  floor 
to  an  intensity  of  1.4  foot-candles.  How  high  should  a  cluster 
of  four  20-c.p.  lamps  be  hung  above  the  space? 

Problem  4-13.  The  average  illumination  on  the  ground  on  a 
moonlight  night  is  .025  foot-candle.  The  moon  is  238,000  miles 
from  the  earth.  What  is  the  candle-power  of  the  moon? 


FIG.  254. — Horizontal  distribution  of 
candle-power  about  a  carbon  incan- 
descent lamp. 


FIG. 


255.— Vertical  distribu- 
tion. 


215.  Distribution  of  Intensity  about  a  Lamp.  Lamps 
do  not  show  the  same  brightness  on  all  sides;  that  is, 
they  have  a  greater  candle-power  in  one  direction  than  in 
another.  It  is  customary  to  obtain  the  candle-power  of 
a  lamp  as  viewed  from  all  positions  and  average  the  results. 
This  average  is  called  the  MEAN  SPHERICAL  CANDLE-POWER 
of  the  lamp. 

When  the  average  candle-power  is  taken  for  points  in  a 
horizontal  plane  only,  this  value  is  called  the  MEAN  HOR- 
IZONTAL CANDLE-POWER.  The  latter  is  the  usual  method 
of  rating  incandescent  lamps. 

Figure  254  shows  the  results  of  some  tests  taken  at 
Pratt  for  the  distribution  of  candle-power  around  a  single- 


368 


ELEMENTS  OF  ELECTRICITY 


loop  carbon-filament   lamp,  in  a  horizontal  plane.     Notice 
that  the  light  is  very  evenly  distributed  in  this  plane. 

Figure  255  shows  the  vertical  distribution  around  the 
same  lamp.  Note  that  the  lamp  has  low  candle-power 
directly  beneath  the  tip. 


Fie.  256. — Distribution  when  fitted  with  special  shade. 

Figure  256  shows  the  effect  on  the  distribution  when  a 
special  shade  is  used  on  the  same  lamp  in  order  to  throw 
the  light  down. 

216.  Measurement  of  Candle-power.  Photometers.  The 
candle-power  of  a  lamp  is  measured  by  comparing  it  with 
another  lamp  of  known  candle-power.  The  comparison 
is  generally  made  by  means  of  an  instrument  called  a 
PHOTOMETER. 


B 

<s 

r\x 

B 

•fc.| 

"Oil  Spot  ft 

r 

1 

I                                 / 

\ 

-i          r 

\ 

J          LJ 

ihli                                 hhlilil 

i 

FIG.  257. — Diagram  of  Bunsen  photometer. 

We  will  take  up  the  simplest  form  of  this  instrument, 
the  Bunsen  Photometer,  shown  diagrammatically  in  Fig. 
257. 


PHOTOMETRY  AND  ELECTRIC  ILLUMINATION    369 

The  standard  lamp  A  and  the  lamp  to  be  tested,  X,  are 
placed  at  the  ends  of  a  track.  Between  the  lamps  is  placed 
a  paper  screen  B,  on  which  is  a  paraffin  oil  spot.  There 
should  be  no  source  of  light  in  the  room  except  X  and 
A.  If  the  screen  is  placed  at  such  a  position  that  it  is 
illuminated  equally  on  both  sides,  the  oil  spot  disappears. 

When  this  position  has  been  found,  let 

R=  distance  of  screen  from  lamp  X. 
S=  il  "  "     A. 

A  =  intensity  of  A  in  candle-power. 
X=  "        X 

Then  the  intensity  of  illumination  on  each  side  of  the  screen 
is  as  follows: 

^ 

-02=  intensity  on  side  facing  A; 

Y" 

**•  i  (  (  C  (  i  -y 

& 

But  intensity  is  equal  on  both  sides,  hence 

A=^ 

S2~R2 
or 

R2 
Y-—  4 

~S2 

Thus  the  intensities  of  the  lamps  are  to  each  other  as  the 
squares  of  their  respective  distances  from  the  screen. 

Problem  6-13.  In  Fig.  257  standard  lamp  A  is  16  c-p., 
»S  =  230  cms.,  72  =  250  cms.  What  is  the  candle-power  of  lamp 
X? 

Problem  7-13.  If  the  distance  from  A  to  the  screen,  Fig.  257, 
is  twice  as  great  as  the  distance  from  X  to  the  screen,  how  does 
the  candle-power  of  A  compare  with  that  of  Xl 

In  practice  the  standard  lamp  is  usually  an  incandescent 
lamp  with  a  carbon  filament,  carefully  standardized  at  a 


370  ELEMENTS  OF  ELECTRICITY 

rated  voltage.  These  standard  lamps  should  not  be  burned 
for  any  length  of  time,  as  they  are  likely  to  change  in  candle- 
power.  A  standard  should  always  be  used  which  is  as 
nearly  as  possible  of  the  same  candle-power  as  the  lamp  to 
be  tested,  and  should  always  be  burned  at  its  rated  voltage. 
Figure  258  represents  a  common  method  of  obtaining 
proper  voltage  regulation.  The  variable  resistance  RI  is 
adjusted  until  the  desired  voltage  is  obtained  across  the 
lamp  X,  and  then  variable  resistance  R2  adjusts  the  voltage 
across  the  standard  lamp  A.  For  this  arrangement,  lamp 
A  must  be  of  lower  rated  voltage  than  lamp  X.  Since 
they  are  in  parallel  across  the  same  mains,  any  fluctuation 


FIG.  258. — Diagram  of  photometer  wiring  for  testing  incandescent  lamps. 

in  the  voltage  during  readings  will  cause  approximately 
proportional  changes  in  candle-power  of  both  lamps  and 
not  affect  the  results. 

The  lamp  X  is  often  rotated  at  about  120  revolutions 
per  minute.  Its  mean  horizontal  candle  power  can  thus 
be  found  from  one  setting  of  the  screen. 

For  obtaining  the  candle  power  of  arc  lamps,  special 
photometric  apparatus  has  to  be  used,  for  a  description  of 
which,  see  "  High  Efficiency  Electrical  Illuminants,"  by 
R.  V.  Hutchinson,  Jr. 

217.  Measurement  of  Intensity  of  Illumination.  The 
Illuminometer.  The  problem  of  computing  or  measuring 
the  intensity  of  illumination  on  any  given  surface  is  very 
complex.  If  only  one  source  of  light  is  illuminating  the 


PHOTOMETRY  AND  ELECTRIC  ILLUMINATION    371 


surface  which  is  perpendicular  to  the  rays  of  light,  a  direct 
application  of  the  law  of  inverse  squares  would  be  all  that 
is  necessary.  But  such  is  never  the  case  in  actual  practice. 
Light  generally  comes  from  several  sources,  either  directly 
or  after  being  reflected  from  other  surfaces. 

Several  instruments  have  been  devised  to  measure  the 
intensity  of  illumination  directly. 

The  portable  photometer  shown  in  Figs.  259  and  260 
is  a  most  convenient  and  satisfactory  device  for  the  meas- 


FIQ.  259. — Side  view  of  Sharp-Millar  photometer. 

urement  of  both  the  intensity  of  light  sources  directly 
in  candle-power,  and  of  intensity  of  illumination  of  sur- 
faces directly  in  foot-candles. 

It  is  called  the  Sharp-Millar  Universal  Photometer  and 
consists  of  a  closed  box  containing  a  standard  lamp  L, 
which  illuminates  an  opaque  screen  S,  and  lights  up  one 
side  of  the  photometer  screen  P. 


FIG.  260.— Top  view  of  Sharp-Millar  photometer. 


The  light  whose  candle-power  is  to  be  measured  is 
placed  at  a  stated  distance  from  the  opening  in  the  tube 
T,  and  illuminates  a  white  matt  surface  placed  in  the  tube 
at  M.  The  light  reflected  from  this  matt  surface  illumi- 
nates the  other  side  of  the  Photometer  Screen  P. 

We  thus  have  a  photometer  screen  illuminated  on 
one  side  by  light  from  a  lamp  of  known  candle-power,  and 
on  the  other  by  light  from  a  lamp  of  unknown  candle- 
power.  We  merely  have  to  get  the  illumination  the  same 


372  ELEMENTS  OF  ELECTRICITY 

on  both  sides  of  the  screen  in  order  to  determine  the  candle- 
power  of  the  unknown  source. 

This  is  done  by  drawing  the  standard  lamp  along  in  the 
box  by  means  of  the  knurled  head  K,  until  the  spot  on 
the  photometer  screen  disappears  as  in  the  Bunsen  photom- 
eter. Instead  of  having  now  to  measure  the  distance 
from  the  lamps  to  the  screen,  a  spot  of  light  on  the  scale 
A  indicates  directly  the  candle-power  of  the  unknown  source. 

If  the  lamp  of  unknown  candle-power  is  not  placed  at 
the  correct  distance  from  the  photometer  box,  of  course 
a  correction  has  to  be  applied  to  the  indicated  candle- 
power  as  follows: 

The  indicated  candle-power  _ 
True  candle-power 

Square  of  standard  distance  of  lamp  from  surface  M 
Square  of  actual  distance  of  lamp  from  surface  M 

When  it  is  desired  to  measure  the  intensity  of  illumi- 
nation on  any  surface,  the  photometer  is  brought  to  that 
place,  an  opaque  screen  like  that  at  S  is  placed  at  the  open- 
ing 0,  and  a  mirror  replaces  the  white  matt  surface  at  M. 
The  illumination  on  the  side  of  the  photometer  screen 
P  toward  lamp  L  is  now  regulated  as  above  by  moving 
the  standard  lamp.  The  scale  now  indicates  the  illumi- 
nation of  the  screen  at  0  in  foot-candles. 

The  standard  lamp  consists  of  a  one  candle-power  tung- 
sten incandescent  lamp  run  on  a  small  storage  battery. 
It  takes  about  .9  ampere  at  4  volts. 

In  order  to  measure  the  candle-power  of  lamps  of  high 
intensity,  a  smoked  glass  absorbing  plate  may  be  placed 
between  the  photometer  screen  and  the  white  matt  surface 
M.  This  cuts  down  the  light  reaching  the  photometer 
screen  P  by  a  known  amount,  so  that  the  indicated  candle- 
power  has  merely  to  be  multiplied  by  some  •  constant  of 
about  10,  or  100,  depending  on  the  absorbing  plate  used. 


PHOTOMETRY  AND  ELECTRIC  ILLUMINATION     373 

When  measuring  very  weak  illumination,  the  absorbing 
plate  is  placed  between  the  screen  S  and  the  photometer 
screen.  The  indicated  scale  reading  must  then  be  divided 
by  10  or  100,  as  the  case  may  be. 

By  means  of  these  absorbing  screens  which  can  quickly 
be  turned  into  place,  intensities  can  be  measured  varying 
from  .004  foot-candle  to  2000  foot-candles. 

The  instrument  surely  has  a  right  to  the  name  of  Universal 
Photometer. 

218.  Arc  Lamps.  When  an  electric  current  is  made 
to  flow  across  a  small  gap  from  the  tip  of  one  carbon  rod 
to  another,  an  electric  "  arc  "  is  formed.  The  carbon 
points  are  heated  to  a  brilliant  incandescence,  gradually 
vaporized  and  burned  up  by  the  oxygen  in  the  air.  The 
electric  current  passes  from  one  tip  to  the  other  in  this 
stream  of  incandescent  carbon  vapor.  If  the  rods  are 
composed  of  pure  carbon,  by  far  the  greater  part  of  the 
light  is  emitted  by  the  glowing  tips  of  the  rods,  and  but 
little  by  the  carbon  vapor.  Fig.  261  shows  the  results 
of  a  test  on  the  distribution  about  a  direct  current  arc 
lamp.  The  light  is  seen  to  have  the  greatest  intensity 
at  an  angle  of  40  to  45  degrees  below  the  horizontal.  Thus 
most  of  the  light  comes  from  the  "  crater  "  in  the  upper 
or  positive  carbon.  The  light  being  thrown  downward 
makes  this  form  of  arc  especially  adapted  to  street  lighting. 
In  an  alternating  current  arc  lamp  the  light  is  sent  out 
in  about  equal  intensity  from  both  carbon  tips  as  shown 
in  Fig.  262.  The  light  emitted  by  carbon  arcs  is  an  intense 
white,  slightly  bluish.  They  have  the  great  advantage  of 
causing  objects  to  appear  in  their  normal  colors.  But  they 
also  have  the  disadvantage  of  any  source  of  light  of  high 
candle-power,  in  that  the  eye  must  be  shaded  from  the 
direct  rays.  When  shades  are  employed  for  this  purpose, 
only  about  half  of  the  total  light  can  be  used. 

If  the  rods  are  formed  of  a  mixture  of  carbon  and  calcium 
fluoride,  the  vapor  given  off  becomes  highly  luminous, 


374 


ELEMENTS  OF  ELECTRICITY 


a  golden  orange  in  color,  and  often  producing  light  of  1000 
mean  spherical  candle-power.  These  so-called  LUMINOUS 
or  FLAMING  ARCS  are  extensively  used  for  display  and 
advertising  purposes.  The  fumes  given  off  are  obnoxious, 
so  that  they  are  rarely  used  for  indoor  lighting.  Further- 
more, the  cost  of  the  special  carbons  and  their  rapid  con- 
sumption has  hindered  their  coming  into  general  use,  in 
spite  of  their  high  efficiency  of  .15  watt  per  candle-power. 


FIG.  261. — Distribution   of  candle- 
power  about  a  D.C.  arc  lamp. 


FIG.  262. — Distribution  of  candle-power 
about  an  A.C.  arc  lamp. 


To  decrease  the  rate  of  consumption  of  ordinary  carbon 
rods,  the  arc  is  sometimes  inclosed  in  a  nearly  air-tight 
glass  globe.  This  excludes  the  oxygen  to  such  an  extent 
that  the  same  pair  of  carbons  lasts  100  hours  instead  of 
only  7  or  8. 

From  70  to  80  volts  can  be  used  to  maintain  a  long  arc 
in  the  inclosed  lamp,  while  only  35  volts  can  be  used  across 
a  short  arc  in  the  open  air.  The  inclosed  arc  can  thus 
be  used  on  a  110-volt  circuit.  These  facts  are  tending  to 
cause  the  inclosed  arcs  to  be  used  in  almost  all  new  instal- 
lations. 


PHOTOMETRY  AND  ELECTRIC  ILLUMINATION     375 

218.  Operation  of  D.C.  Arc  Lamps.  Ballasting  Resist- 
ance. The  arc  between  the  tips  of  carbon  rods  is  naturally 
unstable  and  cannot  be  maintained  alone,  directly  across 
the  supply  mains  of  constant  potential  circuit.  This 
is  evident  if  we  consider  the  characteristics  of  the 
arc. 

Suppose  the  arc  to  be  burning  steadily  across  akconstant 
potential  source  of  supply.  If  the  voltage  decreased  ever 
so  little,  the  current  would  decrease  a  little  and  would  not 
vaporize  the  carbon  so  rapidly.  The  path  of  carbon  vapor 
between  the  tips  would  then  be  less  dense  or  of  smaller 
cross-section;  so  the  resistance  of  it  would  increase.  This 
would  still  further  decrease  the  current,  and  continue  to 
do  so  till  the  arc  was  extinguished. 

Or  suppose  the  voltage  increased  ever  so  little.  The 
current  would  increase  and  cause  a  greater  flow  of  carbon 
vapor  between  the  tips.  This  would  decrease  the  resistance 
of  the  arc  and  allow  still  more  current  to  flow.  This  would 
continue  until  practically  a  short  circuit  was  made. 

All  of  these  changes  would  take  place  too  rapidly  for 
any  regulating  device  to  counteract  them. 

The  instability  of  such  an  arc  is  corrected  as  follows: 

A  BALLASTING  RESISTANCE  R,  Fig.  263,  is  inserted  in 
series  with  the  arc  A ,  across  the  line  of  constant  potential  M. 

The  voltage  across  the  mains  is  then  taken  up  both  in  the 
IR  drop  of  the  ballasting  resistance  and  in  the  voltage  used  in 
maintaining  the  arc.  If  the  voltage  across  the  mains  becomes 
a  little  greater  and  causes  a  greater  current  to  flow  through 
the  lamp,  the  IR  drop  across  the  ballast  becomes  enough 
greater  to  cause  a  slight  decrease  in  the  voltage  across 
the  arc.  Any  decrease  in  current  through  the  lamp  causes 
a  less  IR  drop  across  the  ballast  and  allows  a  greater  part 
of  the  line  voltage  to  come  across  the  arc.  The  ballasting 
resistance  then  opposes  an  increase  or  decrease  of  current 
between  the  carbon  tips.  It  must  always  be  inserted 
when  lamps  are  used  on  constant  potential  lines.  In 


376 


ELEMENTS  OF   ELECTRICITY 


A.C.  lamps  a  choke  coil  may  take  the  place  of  the  ballasting 
resistance.     For  action  of  choke  coil,  see  Chapter  XV. 

REGULATING  COIL.  The  device  for  regulating  the  length 
of  the  arc  is  very  simple  when  the  lamps  are  used  in  paral- 
lel on  a  constant  potential  line.  See  Fig.  264.  In  series 
with  the  ballast  R  and  the  arc  A  is  placed  a  REGULATING 
COIL  C.  The  plunger  P  is  attached  to  a  friction  clutch  F. 
AVhen  no  current  is  running  through  the  lamp  the  plunger 
drops  to  its  lowest  position  and  causes  the  friction  clutch  to 


FIG.  263.— Arc  and  ballast- 
ing resistance. 


FIG.  264. — Arc  with  ballasting  and  regu- 
lating resistances. 


release  the  upper  carbon  and  allow  it  to  drop  until  it 
touches  the  lower  carbon. 

The  two  carbons  are  thus  in  contact  when  the  power  is 
off.  So,  when  the  current  is  turned  on,  there  is  a  complete 
circuit  through  the  lamp.  But  as  the  current  goes  through 
the  coil,  the  plunger  P  is  sucked  up  into  the  coil.  The 
friction  clutch  grips  the  upper  carbon  and  raises  it,  on 
account  of  this  upward  motion  of  the  plunger.  An  arc 
is  thus  struck  between  the  positive  and  negative  carbon 
tips.  These  parts  are  shown  in  Fig.  265. 

As  the  carbons  burn  away  and  increase  the  length  of 
the  gap  between  them,  and  accordingly  increase  the  resist- 


PHOTOMETRY  AND  ELECTRIC  ILLUMINATION    377 

ance   of  the   circuit,  less   current   flows.     This  makes  the 

regulating   coil   weaker   and   allows   the   plunger  to   drop. 

The  dropping  of  the  plunger  releases  the  friction   clutch 

on  the  positive  carbon,  so  that  it  falls 

and    touches    the    lower    carbon   again. 

The    process   of   striking   the    arc   then 

takes  place  as  at  first.     Thus  an  almost 

constant  length    of    arc    is  maintained. 

The    voltage    across    the    arc    can    be 

changed    by    adjusting    the   amount   of 

resistance  in  the  ballast.      The  amount 

of  current  taken  by  the   lamp  can  be 

regulated  by  adjusting  the  distance  the 

plunger  is  allowed  to  move. 

When  arc  lamps  are  operated  in 
series,  a  much  more  complicated  regu- 
lating scheme  is  used.  A  ballast  resist- 
ance, however,  need  not  be  inserted  in 
the  lamps,  because  an  automatic  adjust-  FIG  265.  —  Photograph 

"of  arc   lamp    showing 

ment    is    made    at    the    generator  for      ballasting  and  regulat- 
ing resistances. 

keeping  the  current  output  constant. 

219.  Incandescent  Lamps.  Carbon  Filament.  When 
an  electric  current  of  sufficient  strength  is  sent  through 
a  fine  filament  of  carbon  it  heats  it  to  incandescence  and, 
if  exposed  to  the  oxygen  of  the  air,  the  filament  immediately 
burns  up.  For  this  reason  the  filament  has  to  be  inclosed 
in  a  bulb  from  which  the  air  has  been  exhausted  when  this 
source  of  light  is  to  be  used.  Such  a  lamp  can  be  made 
of  any  medium  candle-power  and  emits  a  soft  yellow  light, 
very  well  suited  to  indoor  illumination. 

The  resistance  of  the  cold  carbon  filament  is  about  twice 
as  high  as  when  incandescent,  though  any  further  increase 
in  temperature  does  not  greatly  tend  to  further  decrease 
the  resistance.  Such  a  lamp  is  rated  as  to  its  voltage, 
mean  horizontal  candle-power,  and  life.  If  a  lamp  is 
burned  at  any  voltage  differing  but  slightly  from  that 


378 


ELEMENTS  OF  ELECTRICITY 


specified  in  the  rating,  a  great  change  in  the  candle-power, 
efficiency  and  life  will  take  place. 

The  following  table  shows  the  effect  on  the  candle-power^ 
efficiency  and  life,  .when  a  lamp  rated  at  110  volts,  16  c.p. 
is  burned  at  either  lower  or  higher  voltages. 


Actual 
Volts. 

Per  Cent 
of  Rated 
Voltage. 

Mean 
Horizontal 
Candle- 
power. 

Percentage 
of  Rated 
C.P. 

Life 
in 
Hours. 

Per  Cent 
of  Rated 
Life. 

Watta 
per 
C.P. 

90 

55 

760 

91 

59 

650 

92 

63 

545 

93 

67 

450 

94 

71 

355 

95 

75 

270 

105. 

96 

12.8 

80 

1,000 

200 

3.86 

106.7 

97 

13.6 

85 

875 

175 

3.72 

107.8 

98 

14.4 

90 

735 

147 

3.60 

108.9 

99 

15.2 

95 

600 

120 

3.50 

110 

100 

16.0 

100 

500 

100 

3.40 

111.1 

101 

17.0 

106 

400 

80 

3.28 

112.2 

102 

17.8 

111 

300 

60 

3.16 

113.3 

103 

18.9 

118 

240 

48 

3.06 

114.4 

104 

19.9 

124 

200 

40 

2.96 

105 

131 

34 

106 

138 

29 

107 

145 

25 

108 

153 

21 

109 

161 

18 

110 

169 

15 

Note  that  a  rise  of  1  per  cent  above  the  rated  voltage 
causes : 

A  rise  of  6  per  cent  in  candle-power. 

A  loss  of  20  per  cent  in  life. 

A  gain  of  3^  per  cent  in  efficiency. 
A  fall  of  1  per  cent  below  rated  voltage  causes: 

A  loss  of  5  per  cent  in  candle-power. 

A  gain  of  20  per  cent  in  life. 

A  loss  of  3  per  cent  in  efficiency. 


PHOTOMETRY  AND  ELECTRIC  ILLUMINATION    379 

This  illustrates  the  necessity  of  avoiding  burning  a  lamp 
much  above  or  below  its  rated  voltage. 

Problem  8-13.  A  3.5-watt  carbon  filament  lamp  which  has 
18  candle-power  at  110  volts,  will  have  what  candle-power  at 
107  volts? 

Problem  9-13.  What  would  be  the  efficiency  of  lamp  in  Prob- 
lem 8  at  107  volts? 

Problem  10-13.  How  much  would  the  life  of  the  lamp  in 
Problem  8  be  affected  by  being  used  on  107  volts? 

Problem  11-13.  What  would  be  the  effect  upon  the  life, 
efficiency  and  candle-power  on  a  3.30  watt,  16  c.p.,  112-volt 
carbon  filament,  if  it  were  burned  on  115  volts? 

The  color  of  the  light  emitted  by  a  carbon  incandescent 
lamp  is  very  pleasing  to  the  eyes,  but  contains  too  much 
yellow  to  give  the  normal  color  to  objects.  In  this  respect 
it  is  inferior  to  the  arc  lamp.  Its  efficiency  is  rarely  better 
than  3.5  watts  per  candle-power  except  in  the  case  of 
the  G.E.M.  lamp,  which  has  an  efficiency  of  about  2.5  watts 
per  candle-power.  The  filament  in  the  latter  is  composed 
of  carbon  which  has  been  "  metallized  "  by  a  special  firing 
process. 

220.  Tantalum  and  Tungsten  Lamps.  Incandescent 
lamps  have  lately  come  into  use,  in  which  the  filaments 
are  of  metal,  either  tantalum  or  tungsten.  These  filaments 
have  to  be  made  very  fine  and  much  longer  than  a  carbon 
filament,  in  order  to  get  the  resistance  necessary  on  a  110- 
volt  circuit.  The  arrangement  of  filament  is  illustrated 
in  Fig.  266.  The  efficiency  of  a  TANTALUM  lamp  is  about 
2  watts  per  candle-power;  and  the  horizontal  distribution 
is  remarkably  uniform.  The  life  on  a  D.C.  circuit  is  from 
700  to  900  hours,  but  falls  to  about  half  that  on  an  A.C. 
circuit,  due  to  a  peculiar  granulating  effect  which  the  alter- 
nating current  has  on  the  filament. 

The  TUNGSTEN  lamp  has  an  efficiency  of  from  1  to  1.5 
watts  per  candle-power,  a  life  of  1000  hours,  and  is  unaf- 


380 


ELEMENTS  OF  ELECTRICITY 


fected  by  an  alternating  current.  The  main  difficulty 
in  the  use  of  both  tantalum  and  tungsten  lamps  is  the 
fact  that  the  long  slender  filaments  are  so  liable  to  breakage. 
This  is  especially  true  of  the  tungsten. 

The  filament  of  each  becomes  more  or  less  plastic  when 
burning,    and    sags   to  such  an  extent   when    placed  in  a 

horizontal  position,  that  parts  of 
it  easily  form  short  circuits.  The 
extreme  high  efficiency  and  long 
life  (barring  accidents)  of  the 
tungsten  has  made  it  a  favorite 
for  indoor  lighting  and  it  is  rapidly 
displacing  the  carbon  filament 
lamps.  The  color  also  is  nearer 
white  and  gives  more  nearly  norm'al 
color  values  to  illuminated  sur- 
faces. The  General  Electric  Com- 
pany have  recently  announced  a 
tungsten  lamp,  the  filament  of 
which  is  drawn  and  exceedingly 
tough.1 

221.  The  Nernst  Lamp.  In  the 
Nernst  lamp,  use  is  made  of  the 
fact  that  many  non-conducting  sub- 
stances become  conductors  when  heated  to  incandescence.  A 
small  pencil-like  glower  composed  of  about  the  same  mate- 
rials as  the  Welsbach  gas  mantle  is  first  heated  by  an  aux- 
iliary heating  device  to  make  it  a  conductor.  A  current 
can  then  be  sent  through  it,  which  causes  it  to  glow  with 
an  intense  white  light.  The  glower,  being  in  the  bottom  of 
the  lamp,  throws  most  of  the  light  down.  It  is  generally 
surrounded  by  a  frosted  globe  on  account  of  its  high  candle- 

1  The  reason  for  the  higher  efficiency  of  metallized  carbon,  tantalum 
and  tungsten  lamps,  lies  in  the  fact  that  they  operate  at  higher  tem- 
peratures and  thus  give  out  a  greater  percentage  of  the  energy  in 
light  waves  and  a  less  percentage  in  heat  waves. 


FIG.  266. — Arrangement  of  long 
filament  in  tantalum  lamp. 


PHOTOMETRY  AND  ELECTRIC  ILLUMINTAION     381 


power.  The  air  does  not  have  to  be  excluded  from  the 
glower  as  from  an  incandescent  filament.  Fig.  267  repre- 
sents the  structure  of  the  lamp  diagrammatically. 

The  glower  G  is  connected  in  series  with  the  coil  M  and 
the  ballast  R.  The  heating  coils  C  are  at  first  connected 
in  parallel  with  the  glower. 

When  the  power  is  turned  on,  the  current  will  not  go 
through  the  glower,  as  it  is  a  non-conductor  when  cold. 
The  coil  M  is  therefore  not  energized,  the  current  taking 
the   parallel   path  through    the 
heating  coil  C  by  way  of   the 
spring  steel  arms  A. 

In  a  half  minute,  the  heating 
coil  C  has  raised  the  temperature 
of  the  glower  high  enough  to 
cause  it  to  become  a  conductor. 
The  current  now  flows  through 
the  path  containing  the  glower 
energizing  the  coil  M ,  the  pole 
piece  of  which  P,  attracts  the 
arms  A,  and  breaks  the  heating 

nr\\]     niTnnif         TVio  -nr 
COll    Circuit.       ine  W 

now  going  through  the  glower, 

heats  it  to  incandescence.  When  very  high  candle- 
power  is  desired,  several  glowers  are  connected  in  multiple. 
The  lamp  has  excellent  color  qualities;  and  an  efficiency 
of  about  2  watts  per  candle-power.  The  life  of  a  glower, 
which  can  be  replaced,  is  about  700  hours. 

The  complicated  mechanism  and  high  cost  of  glowers 
do  not  allow  the  Nernst  lamp  to  compete  in  this  country 
with  the  highly  efficient  filament  lamps  described  above. 

222.  Mercury  Arc  Lamp.  The  mercury  arc  lamp, 
commonly  known  as  the  Cooper  Hewitt  lamp,  makes  use 
of  the  property  of  luminescence  of  metallic  vapor.  The 
mercury  is  held  in  the  lower  end,  M,  Fig.  268,  of  a  vacuum 
tube  20  to  50  inches  long. 


on  vron+     FIG.  267. — Diagram   of  electric  cir- 
cuits within  a  Nernst  lamp. 


382 


ELEMENTS  OF   ELECTRICITY 


Since  current  will  not  flow  through  the  tube  from  B  to 
A  unless  there  is  a  stream  of  vapor  to  carry  it,  a  starting 
device  must  be  used. 

Sometimes  the  tube  is  merely  tilted  back  into  the  horizon- 
tal position  so  that  a  thread  of  mercury  flows  out  of  M 
and  completes  the  circuit.  It  is  then  allowed  to  return 
to  a  position  at  a  slight  angle  to  the  horizontal,  so  that  the 
mercury  will  flow  back  into  the  end  as  the  vapor  condenses. 
Generally  an  automatic  starting  device  is  used.  It  con- 
sists of  a  coil,  L,  of  high  inductance,  through  which  the  cur- 
rent goes  at  first.  But  this 
circuit  is  suddenly  broken 
by  an  ingenious  auto- 
matic mercury  contact 
breaker  D.  The  high  self- 
inductance  of  the  circuit 
causes  a  spark  to  jump 
across  from  G  to  A.  This 
spark  charges  the  mercury 
electrically  and  sets  up 
enough  vapor  in  the  tube 
for  the  current  to  pass 
from  B  to  A.  The  tube 
now  glows  with  a  light  com- 
posed entirely  of  green,  blue  and  yellow  waves.  This 
gives  a  peculiar  color  effect  to  most  objects,  especially 
those  which  usually  reflect  red.  The  light  has,  however, 
a  very  penetrating  effect  and  brings  out  the  shape  and 
contour  of  objects  very  distinctly.  Its  intrinsic  brilliancy 
is  low,  being  only  about  17  c.p.  per  sq.in.  of  tube. 
For  this  reason,  it  is  used  in  drafting  rooms  and  wherever 
a  light  is  needed  to  make  small  details  stand  out  clearly. 

A  50-inch  tube  has  a  hemispherical  candle-power  of 
about  800  c.p.  and  takes  3.5  amperes  at  110  volts,  or  .48 
watt  per  c.p.  The  life  is  about  1000  hours,  being  limited 
by  the  loss  of  the  vacuum. 


FIG.  268. — Diagram  of  electric  circuits  in 
a  mercury  vapor  lamp. 


PHOTOMETRY  AND  ELECTRIC  ILLUMINATION     383 

Since  the  arc  is  unstable  below  a  certain  current  strength, 
and  the  lamp  has  the  best  efficiency  at  the  critical  value 
of  current,  a  ballasting  resistance  R  and  inductance  S 
are  used  in  series  with  the  tube  to  counteract  the  effects 
of  any  variation  in  the  voltage  and  keep  the  current  above 
critical  value. 

223.  The  Moore  Tube.  A  high  voltage  will  force  a 
current  of  electricity  through  a  tube  which  contains. gas 
at  a  very  high  vacuum,  and  cause  the  gas  to  glow  with  a 
soft  luminescence.  The  color  depends  upon  the  gas  used, 
carbon  dioxide  producing  a  white  diffused  daylight,  nitrogen 
an  orange  light,  etc. 

The  Moore  Tube,  making  use  of  these  characteristics, 
consists  of  a  long  high-vacuum  tube,  to  the  terminals  of 
which  is  supplied  an  alternating  current  at  high  pressure. 
A  special  transformer  is  employed  to  attain  the  necessary 
high  voltage. 

As  in  operation,  the  gas  is  gradually  precipitated  on  the 
glass  wall  to  the  tube,  the  vacuum  continues  to  grow 
greater.  Thus  it  is  necessary  from  time  to  time  to  admit 
gas  into  the  tube. 

Mr.  Moore  has  devised  a  very  ingenious  valve  for  this 
purpose  which  operates  automatically  and  admits  the 
required  amount  of  gas  at  the  proper  time. 

The  intensity  is  only  about  .65  c.p.  per  sq.in.  This 
requires  that  long  tubes  be  installed,  to  give  sufficient 
illumination  for  large  rooms.  But  the  effect  is  very 
pleasing,  approaching,  as  it  does,  so  closely  to  diffused 
daylight.  The  efficiency  compares  favorably  with  the  best 
carbon  filament  incandescent  lamps.  Fig.  355a  shows  this 
light  installed  in  a  silk  mill,  where  it  finds  special  applica- 
tion in  bringing  out  the  true  color  values  of  the  manufac- 
tured product. 


384  ELEMENTS  OF  ELECTRICITY 


SUMMARY  OF  CHAPTER  XIII 

The  merits  of  any  scheme  of  illumination  should  be  judged 
by  its  effect  upon  the  eye.  Colors  should  appear  as  they  do 
in  daylight  and  objects  should  be  evenly  illuminated  in  such 
a  way  as  to  show  their  outlines. 

NATURE  OF  LIGHT.  Light  is  caused  by  waves  in  the 
ether;  a  definite  color  effect  being  produced  by  a  wave  of 
a  definite  length.  The  violence  of  the  wave  determines  the 
intensity  of  the  light.  An  object  has  the  color  effect  of  the 
waves  it  reflects. 

PHOTOMETRIC  UNITS.  The  unit  of  intensity  of  a  source 
of  light  is  the  candle-power,  which  is  the  intensity  of  a  stand- 
ard candle. 

The  unit  of  intensity  of  illumination  of  a  surface  is  the 
foot-candle,  which  is  the  intensity  of  illumination  on  a  sur- 
face at  right  angles  to  rays  of  light  and  at  one  foot  distance 
from  a  light  of  one  candle-power. 

Intensity  of  illumination  on  any  surface  at  right  angles 
to  rays  of  light  can  be  found  from  the  equation  : 


where  E=  illumination  of  surface  in  foot-candles; 

I  =  intensity  of  source  of  light  in  candle-power; 
d=  distance  of  surface  from  source  hi  feet. 

LAW  OF  INVERSE  SQUARES.  The  intensity  of  illumi- 
nation on  any  surface  at  right  angles  to  light  rays,  is  inversely 
proportional  to  the  square  of  its  distance  from  the  source 
of  light. 

BUNSEN  PHOTOMETER.  A  light  of  unknown  candle- 
power  is  placed  on  one  side  of  a  screen  of  paper  with  an  oil 
spot  on  it.  A  light  of  known  candle-power  is  placed  on  the 
other  side,  and  screen  adjusted  so  that  the  oil  spot  disappears. 
The  candle-powers  of  the  two  lights  are  to  each  other  as  the 
squares  of  their  respective  distances  from  the  screen. 

The  Sharp-Millar  Universal  Photometer  is  a  very  satis- 
factory direct  reading  Photometer  and  Illuminometer. 

ARC  LAMPS,  B.C.  Light  comes  mostly  from  the  crater  in 
upper  carbon  and  is  thrown  down,  unless  carbons  are  impreg- 
nated with  some  metallic  salt,  in  which  case  the  arc  itself 
is  luminous. 


PHOTOMETRY  AND  ELECTRIC  ILLUMINATION    385 

Will  not  operate  satisfactorily  on  constant  potential  lines 
without  ballasting  resistance. 

Arc  is  struck  and  maintained  by  regulating  coil  generally 
in  series  with  arc. 

High  in  efficiency  but  of  too  great  intensity  for  indoor 
lighting  unless  screened.  Colors  appear  nearly  the  same 
as  in  daylight. 

INCANDESCENT  LAMPS.  Carbon  filament;  pleasant  yel- 
low light,  low  efficiency  but  of  convenient  candle-power. 
Small  variation  in  voltage  across  terminals  makes  great  varia- 
tions in  candle-power  and  life.  Filament  must  be  in  vacuum. 

Tungsten  Lamp;  whiter  light  than  carbon  filament,  high 
efficiency,  but  the  filament  is  fragile. 

Nernst  Lamp;  the  glower  is  heated  to  incandescence  in 
the  air.  As  glower  is  a  non-conductor  when  cold,  a  heating 
device  is  added  to  the  lamp.  About  the  same  efficiency  as 
carbon  filament  lamps. 

MERCURY  ARC  LAMP.  Luminous  mercury  vapor  in 
vacuum  tube.  Lacks  red  rays  so  that  colors  do  not  appear 
in  true  values,  but  has  a  very  penetrating  effect,  and  brings 
out  details.  Highly  efficient. 

MOORE  TUBE.  Luminescent  gases  in  high  vacuum 
tube.  Requires  high  voltage.  Can  use  any  gas;  carbon- 
dioxide  producing  a  soft,  white  light  which  gives  colors  their 
true  value.  Of  lower  intrinsic  candle-power  and  very  pleas- 
ing to  the  eye.  About  the  efficiency  of  carbon  filament  lamps. 


PROBLEMS    ON   CHAPTER  XIII 

12-13.  In  a  photometer  arranged  as  in  Fig.  257,  the  standard 
lamp  is  134  cms.  from  the  screen  and  the  test  lamp  is  158  cms. 
from  the  screen  when  balanced.  If  the  standard  lamp  is  20  c.p., 
what  candle-power  is  the  test  lamp? 

13-13.  What  is  the  foot-candle  illumination  on  each  side  of  the 
photometer  screen  in  Problem  12? 

14-13.  In  using  a  Sharp-Millar  photometer  to  measure  the 
candle-power  of  an  arc  lamp,  the  opening  was  held  140  ft.  from 
the  light.  The  photometer  had  been  standardized  for  12  ft., 
and  registered  7.2  c.p.  What  was  the  true  candle-power  of  the 
arc  lamp? 


386 


ELEMENTS  OF  ELECTRICITY 


15-13.  What  foot-candle  illumination  would  be  registered  at 
the  place  where  the  photometer  was  read  in  Problem  14,  assuming 
all  the  light  came  from  the  arc  lamp? 

16-13.  If  a  standard  lamp  of  25  c.p.  were  used  in  photom- 
eter of  Problem  12,  how  far  from  screen  would  it  have  to  be 
placed  in  order  to  balance  test  lamp? 

16-13.  Each  lamp  in  Fig.  187  is  carbon  filament,  32  c.p.  at 
110  volts.  What  is  the  candle-power  of  each  as  located  in  Fig. 
187? 

17-13.  Five  16-c.p.,  110-volt  carbon  filament  lamps  are  in 
series  in  a  street  car  which  is  the  only  one  on  the  line,  and  is  tak- 
ing 75  amperes.  Generator  voltage  is  550  volts;  trolley  is  No. 
00  hard  drawn  copper ;  track  resistance  is  .03  ohm  per  mile ;  car 
is  1  mile  from  station.  What  is  the  candle  power  of  each  lamp  in 
car? 

18-13.  In  a  room  lighted  by  one  32-c.p.  lamp  hung  8  ft.  from 
the  floor,  what  would  be  the  illumination  on  the  floor  directly 
under  the  lamp? 

19-13.  What  would  be  the  illumination  on  the  floor  in  Problem 
18,  4  ft.  from  a  point  directly  under  lamp?  See  footnote  to  page 
366. 

20-13.  If  the  room  in  Problem  18  had  two  16-c.p.  lamps  hung 
8  ft.  from  the  floor  and  10  ft.  apart,  what  would  be  the  illumi- 
nation on  the  floor  directly  under  each  lamp? 

21-13.  What  would  be  the  illumination  on  the  floor  in  Prob- 
lem 20,  at  a  spot  directly 
under  a  point  half  way  be- 
tween the  two  lamps? 

22-13.  In  a  20X30  ft.  room 
there  are  four  20-c.p.  incan- 
descent lamps  symmetrically 
arranged  as  indicated  in  Fig. 
270,  5  ft.  from  long  side  and 
10  ft.  from  short  side.  Sup- 
pose floor  to  be  divided  into 
5-ft.  squares.  Find  the  illumi- 
nation at  corners  of  squares  marked  1,  2,  3,  4.  Lamps  are  each  8  ft. 
above  floor.  Assume  no  reflection  from  side  walls  or  ceiling. 

23-13.  Find  illumination  in  room  of  Problem  22  at  points 
marked  5,  6,  7  and  8. 




—  < 

^  

)  



1 

8 

4        7 



.  —  ^ 

>  — 

-£ 

>  

3  ? 

5 

6 

FIG.  270. 


CHAPTER  XIV 
ELECTRICAL    MEASURING    INSTRUMENTS 

Galvanometer;  D'Arsonval  and  Thomson  Types — Deflecting  Force; 
Control;  Damping — Sensibility — Shunts;  Ayrtori  Universal  Shunt 
— Series  Resistance:  Megohm — Ballistic  Galvanometer — Thermal 
Effects — Ammeters;  Resistance  of — Types:  Solenoidal,  Hot- Wire.; 
Permanent  Magnets,  Two  Coil  or  Electro-Dynamometer — Volt- 
meters; Resistance  of — Types:  Solenoidal,  Hot-Wire,  Permanent 
Magnet,  Two  Coil,  Electro-Static — Wattmeter — Construction 
of  Weston  Type — Thomson  Integrating  Wattmeter  or  Watt- 
hour  Meter — Potentiometer;  Theory,  Construction  and  Use- 
Voltameter. 

IN  the  following  detailed  description  of  electrical  meas- 
uring instruments,  no  attempt  has  been  made  to  include 
all  electrical  instruments.  It  is  intended  to  study  those 
types  only  which  are  in  general  use  in  this  country,  in 
operating  stations,  laboratories,  and  testing  departments 
of  manufacturing  plants. 

224.  Galvanometer.  The  oldest  instrument  for  meas- 
uring current  and  voltage  is  the  galvanometer.  Most  of 
our  modern  ammeters  and  voltmeters  are  merely  adaptations 
of  one  form  or  other  of  this  instrument. 

In  its  earliest  form,  the  galvanometer  consisted  of  a  com- 
pass needle  suspended  in  the  center  of  a  coil.  Fig.  271 
represents  such  an  instrument,  called  a  tangent  galvanom- 
eter. This  instrument  is  so  called  because  the  current 
flowing  in  the  coil  is  proportional  to  the  tangent  of  the 
angle  through  which  the  needle  is  deflected. 

If  instead  of  reading  the  angle  through  which  the  needle 
swings,  the  coil  is  turned  through  an  angle  sufficient  to 

387 


388 


ELEMENTS  OF  ELECTRICITY 


bring  the  needle  back  to  zero,  the  instrument  becomes 
a  SINE  GALVANOMETER.  The  current  flowing  in  the  coil 
is  then  proportional  to  the  sine  of  the  angle  through  which 
the  coil  has  to  be  turned.  Such  instruments  even  when 
fitted  with  a  mirror  and  telescope  were  not  very  sensitive, 
and  have  only  a  historical  interest. 

225.  Thomson  Astatic  Galvanometer.  Sir  William 
Thomson  (Lord  Kelvin)  made  many  remarkable  improve- 
ments on  the  above  galvanometers  and  succeeded  in  pro- 


FIG.  271. — Tangent  galvanometer.        FIG.  272. — Thomson  astatic  galvanometer. 

ducing  an  instrument  of  high  sensibility.  Fig.  272 
represents  a  typical  galvanometer  of  the  Thomson  type. 

The  two  coils  I  and  II  are  wound  in  opposite  directions. 
When  a  current  is  sent  through  them,  they  are  magnetized 
in  opposite  directions.  This  allows  an  arrangement  of  the 
needles  called  ASTATIC. 

The  needles  within  coils  I  and  II  arc  joined  together  by 
a  stiff  wire  but  in  reversed  positions,  as  in  Fig.  273.  If 
they  are  of  nearly  equal  strength  they  nearly  neutralize 
one  another.  They  will  then  have  but  little  tendency  to 
take  a  definite  position  in  the  earth's  field,  since  they 
are  acted  upon  by  it  to  nearly  equal  degrees  in  opposite 
directions.  In  order  that  such  an  arrangement  of  needles 


ELECTRICAL  MEASURING  INSTRUMENTS         389 

may  return  to  a  definite  zero  point,  a  controlling  magnet, 
NSj  Fig.  272,  is  placed  above  or  below  it.  As  this  magnet 
is  nearer  to  one  needle  than  to  the  other,  it  causes  the 
combination  to  take  a  definite  position.  The  controlling 
force  of  this  magnet  can  be  made  as  small  as  desired.  Thus 
we  have  strong  magnetic  needles  in  a  very  weak  controll- 
ing field.  An  exceedingly  small  current  in  the  coils  then 
exerts  a  torque  sufficient  to  cause  the  magnets  to  turn 
against  this  controlling  field.  Gal- 
vanometers of  this  type  have  been 
made  so  sensitive  that  one  fifty- 
billionth  part  of  an  ampere  would 
cause  a  deflection  of  one  scale  division. 
This  was  the  type  used  for  receiving 
messages  sent  through  the  Atlantic 
cable. 

But    the    Thomson    galvanometer     PIQ    273  _Astatic 
cannot  be  made  to  return  quickly  to        SgfSS^  two  mag" 
zero,  nor  can  it  be  kept  from  swinging 

back  and  forth  for  a  long  time  after  each  deflection.  The 
means  by  which  the  moving  part  of  an  instrument  is 
brought  back  to  zero  is  called  its  CONTROL. 

When  the  moving  part  of  an  instrument  is  kept  from 
swinging  back  and  forth,  it  is  said  to  be  DAMPED,  and  the 
instrument  is  said  to  be  DEAD-BEAT. 

Thomson  galvanometers  can  be  damped  by  mechanical 
means  only,  such  as  a  vane  attached  to  the  suspension  fiber 
which  increases  the  air  resistance  to  turning.  For  this  rea- 
son another  form  of  galvanometer  is  in  much  more  general 
use,  called  after  its  inventor  the  D'Arsonval. 

226.  D'Arsonval  Galvanometer.  Fig.  274  represents  the 
moving  part  of  an  ammeter  which  is  constructed  on  the 
D'Arsonval  principle.  It  contains  the  working  principles 
of  most  of  our  modern  ammeters  and  voltmeters,  and  is 
built  on  the  reverse  principle  of  the  Thomson  type. 
Instead  of  a  stationary  coil  and  moving  magnets,  this  type 


ar- 


390  ELEMENTS  OF  ELECTRICITY 

consists  of  a  permanent  stationary  magnet,  and  a  moving 
coil.  The  soft  iron  core  is  supported  rigidly  between  poles, 
in  order  to  concentrate  the  magnetic  lines  of  force.  The  coil 
swings  freely  around  this  core.  Owing  to  its  strong  mag- 
netic field,  this  type  is  peculiarly  free  from  ordinary  mag- 
netic disturbances  which  greatly  affect  the  Thomson  type. 

The  Deflecting  force  is  the  motor  action  of  a  current  flowing 
in  the  coil  as  explained  in  Chapter  VIII. 

The  Control  is  effected  by  means  of  the  torsional  elasticity 
of  the  suspension  wire. 


FIG.  274. — Ammeter  of  D'Arsonval  type. 

The  Damping  is  very  easily  effected  by  means  of  the 
induced  currents  set  up  in  a  short-circuited  coil  wound  on 
the  moving  coil,  as  explained  in  Chapter  VIII.  Galvanom- 
eters of  this  type  are  made  of  such  a  sensitiveness  that 
one-hundred-millionth  part  of  an  ampere  will  cause  a 
deflection  of  one  scale  division.  Through  small  ranges, 
the  deflections  of  this  galvanometer  are  proportional  to  the 
current  flowing  in  the  coil.  And  since  the  resistance  of 
the  coil  is  constant,  the  voltage  must  vary  as  the  current. 
Therefore  the  deflections  are  also  proportional  to  the  voltages 
across  the  coil. 


ELECTRICAL  MEASURING  INSTRUMENTS         391 

227.  Shunts.  In  order  that  a  galvanometer  may  be 
used  throughout  a  wide  range  of  current  measurements, 
shunts  are  sometimes  placed  around  the  instrument  so 
that  only  a  small  fraction  of  the  current  in  the  line  passes 
through  the  moving  coil. 

Suppose  that  it  is   desired  to   measure  the   current   Im 
flowing  in  the  main  line  in  Fig.  275,  but  that  the  galvanom- 
eter  G    cannot    safely  carry  all    this 
current.     A  shunt,  S,  might  be  placed 
across  the  terminals  of  the  galvanom- 
eter,  which  would  conduct  a  certain 
proportion  of  the  current  around  the 
galvanometer.     But  in  order  to  know  FIG.  275.— Use  Of  shunt  os> 

with  galvanometer  (G). 

how  much   current   is   flowing  in  the 

main  line,  it  is  necessary  to  know  what  fraction  of  the 
total  current  the  galvanometer  carries  and  what  fraction 
the  shunt  carries. 

Let  S=  resistance  of  shunt  in  ohms; 

G  =  galvanometer  in  ohms ; 

Im  =  current  in   main  line ; 

7S=         "          shunt  iS; 

Ig=  galvanometer  G. 

The  circuit  through  S  is  in  parallel  with  the  circuit  through 
G.  Then 


(Current    in    parallel    combination    equals    the    sum    of 
currents  in  each  branch.) 

The  voltage  across  the  Shunt  =I8S     .     (IR) 

11      Galvanometer  =IgG    .     (IR) 
But  voltages  across  parallel  circuits  are  equal. 
Therefore  I0G=ISS, 

or 

(2)  /.=^. 


392  ELEMENTS  OF  ELECTRICITY 

The  relation  of  the  current  in  Galvanometer  (lg)  to  current 
in  main  line  (7m),  may  be  expressed  by  the  fraction  -=£. 

1m 

But  from  (1) 

7  J 

(3) 


•*•  m       *•  g  ~T~  *8 

Substituting   (2)   in   (3), 

Iff  IgS  S 


1m  r         JJG          IgS+IgG  Iff(S+G)  S+G' 

lg+  s         s 

Thus 


(4) 


S+G' 


The  current  through  the  galvanometer  holds  the  same 
relation  to  the  current  in  the  main  line,  that  the  resist- 
ance of  the  shunt  does  to  the  resistance  of  Shunt  plus 
resistance  of  Galvanometer. 

Example.  Assume  the  resistance  of  the  galvanometer,  Fig. 
275,  to  be  2500  ohms  and  the  shunt  to  be  500  ohms.  What  frac- 
tion of  the  current  in  the  main  line  goes  through  the  galvanom- 
eter? 

G  =  2500, 

'  S=  500, 

I          S  500  I 


ImS+G     2500  +  500     6* 

Thus  £  of  the  current  in  the  main  line  passes  through  the  galvanom- 
eter. 

The  same  equation  can  be  used  to  compute  the  amount  of 
resistance  with  which  a  galvanometer  must  be  shunted  in  order 
that  a  given  fraction  of  the  current  in  the  main  line  may  pass 
through  the  galvanometer, 


ELECTRICAL  MEASURING  INSTRUMENTS         393 

Example.  Suppose  it  is  desired  to  shunt  a  galvanometer  in 
above  example  so  that  one-tenth  of  the  current  in  the  main  line 
shall  pass  through  it.  What  resistance  must  the  shunt  be? 


S  +  2500     10 
10S  =  3+2500; 

£  =—    -  =  277.8  ohms. 


Problem  1-14.  A  galvanometer  of  1000  ohms  resistance  has 
a  shunt  of  100  ohms  resistance.  What  part  of  main  current  flows 
through  galvanometer? 

Problem  2-14.  It  is  desired  to  shunt  galvanometer  in  Problem 
1-14  so  that  but  -^  the  main  current  shall  pass  through  it. 
Of  what  resistance  must  the  shunt  be? 

Problem  3-14.  A  galvanometer  of  2000  ohms  resistance  is 
shunted  with  a  5.00-ohm  shunt.  On  a  certain  circuit  it  gives 
a  deflection  of  1.4  scale  divisions  when  thus  shunted.  What 
deflection  would  it  give  on  same  circuit  if  unshunted? 

Problem  4-14.  What  resistance  shunt  would  be  required  to 
cause  a  deflection  of  14  scale  divisions  of  galvanometer  in  Prob- 
lem 3-14  on  same  circuit? 

228.  .Ayrton  Universal  Shunt.  Fig.  276  represents  a 
shunt  arrangement  by  means  of  which  several  shunt  values 
can  be  obtained  throughout  a  wide  range,  depending  on  the 
position  of  the  arm  K.  As  this  arm  is  swung  from  right 
fo  left,  it  lowers,  by  some  decimal  fraction,  the  amount  of 
current  going  through  the  galvanometer.  Since  this 
fraction  is  always  the  same  regardless  of  the  resistance  of 
the  galvanometer,  this  arrangement  is  called  a  UNIVERSAL 
SHUNT. 

These  fractions  do  not  represent  the  ratio  of  the  current 
in  galvanometer  to  the  current  in  the  main  line.  They 
merely  denote  ratio  of  the  current  in  galvanometer  with 


394 


ELEMENTS  OF  ELECTRICITY 


arm  in  given  position,  to  current  in  galvanometer  with 
arm  at  A. 

Thus  when  the  arm  is  at  5,  ^  as  much  current  flows 
through  the  galvanometer  as  when  it  is  at  A,  not  -^ 
of  the  current  in  the  main  line.  The  fraction  of  the  current 
that  flows  through  the  galvanometer  when  the  arm  is  at 
A  depends  upon  the  resistance  of  the  galvanometer,  and 
is,  of  course,  different  in  connection  with  different  galvanom- 
eters. 

But,  regardless  of  the  resistance  of  the  galvanometer, 
when  the  arm  is  at  B,  C,  or  D,  -^  T-^  or  J-Q^  as  much 


FIG.  276. — Diagram  of  Ayrton  universal  shunt. 

current  flows  through  the  galvanometer  as  when  the  arm 
is  at  A.  The  term  UNIVERSAL  is  therefore  a  little  broad. 
The  great  advantage  of  this  shunt  is  the  ease  with  which 
the  same  galvanometer  can  be  used  to  measure  a  wide 
range  of  currents  when  equipped  with  it. 

It  is  necessary,  however,  to  prove  that  the  ratios  are 
true  as  stated,  regardless  of  the  resistance  of  the  galvanom- 
eter. Note  first  that  as  the  arm  moves  from  right  to 
left,  it  cuts  resistance  out  of  the  shunt  and  adds  it  to  the 
galvanometer  circuit. 


ELECTRICAL  MEASURING  INSTRUMENTS         395 

Let  (7=resistance  of  galvanometer  circuit, 

R=  resistance  of  EA. 

Then  BA  is  made   -f^  R, 
and          CA       "         -ffc  R, 

DA      "          9 


Let  I  g=  current  in  galvanometer, 

Im=  current  in  line. 
S  =  resistance  of  shunt. 

/          S 
Then  7^  ^  g^_/^>  as  m  previous  paragraph. 

'm       o  +tr 

When  arm  is  at  A 

I,         S  R 


Im    S+G     R+CT 

When  arm  is  at  B 

I,        S, 


But  £) 

and  GI 

Thus 

LL=     ^i  -IR         =1  (   R  \ 

Im    Si+Gi      .IR+(G+.9R)     1Q\R+GJ' 

This  ratio  is  ^  of  that  when  the  arm  was  at  A. 

This  shunt  works  especially  well  on  ballistic  galvanom- 
eters, and  is  in  common  use  in  capacity  tests. 

229.  Sensibility  of  Galvanometers.  Working  Constant. 
As  there  are  several  uses  to  which  galvanometers  are  put, 
so  there  are  several  ways  of  stating  their  sensibility. 

If  the  galvanometer  is  to  measure  current  or  quantity 
of  electricity,  the  sensibility  is  rated  as  the  fractional  part 
of  an  ampere  or  coulomb  that  will  produce  a  deflection 
of  one  scale  division.  Thus  it  may  be  stated  of  a  certain 
galvanometer  that  its  sensibility  is  .000004  ampere.  By 


396  ELEMENTS  OF   ELECTRICITY 

this  would  be  meant  that  .000004  of  an  ampere  would 
cause  a  deflection  of  one  scale  division.  A  galvanometer 
of  a  sensibility  of  .0000004  ampere  would  be  10  times  as 
sensitive  as  the  first  instrument  because  it  would  require 
but  ^  as  much  current  to  cause  the  same  deflection. 
The  smaller  the  current  to  produce  a  deflection  of  one  scale 
division,  the  greater  the  sensibility. 

Again,  if  the  galvanometer  is  to  be  used  to  measure 
insulation  or  high  resistance,  as  described  in  Chapter  V, 
the  sensibility  may  be  expressed  as  the  number  of  megohms 
which  must  be  connected  in  series  with  it  across  one  volt 
pressure  to  produce  a  deflection  of  one  scale  division. 

Thus  a  galvanometer  of  8000  megohms  sensibility, 
means  a  galvanometer  which  will  give  a  deflection  of  one 
scale  division  when  8000  megohms  resistance  are  in  series 
with  it  across  one  volt.  If  another  instrument  has  a  sen- 
sibility of  16,000  megohms,  it  would  be  twice  as  sensitive 
as  the  first.  The  greater  the  number  of  megohms  through 
which  one  volt  pressure  can  produce  a  deflection  of  one 
scale  division,  the  greater  the  sensibility. 

In  purchasing  a  galvanometer,  much  is  learned  about 
its  characteristics  from  a  statement  of  its  sensibility. 

In  using  a  galvanometer,  however,  it  is  of  greater  im- 
portance to  know  the  WORKING  CONSTANT  of  the  instrument. 
The  term  "  working  constant  "  refers  merely  to  a  local 
''  set-up  "  of  the  instrument  and  not  to  standard  condi- 
tions as  in  the  case  of  the  term  "  sensibility."  For  instance, 
the  sensibility  of  a  galvanometer  may  be  1000  megohms., 
while  its  working  constant  for  a  certain  set-up  may  be  100,000 
megohms.  That  is,  the  galvanometer  may  be  on  a  100- 
volt  line  and  give  a  deflection  of  one  scale  division 
when  100,000  megohms  are  in  series  with  it.  Know- 
ing the  "  working  constant  "  it  is  possible  to  compute  the 
resistance  that  is  in  series  with  a  galvanometer  at  any 
given  time,  by  noting  the  deflection.  Thus  if  the  above 
galvanometer  gave  a  deflection  of  4,  it  would  indicate 


ELECTRICAL  MEASURING  INSTRUMENTS         397 

1 00  000 
that  only  -  -  or   25,000   megohms   must  be  in  series 

with  it  at  that  time.  For  a  complete  discussion  of  the 
method  for  finding  the  working  constant  of  a  galvanometer 
as  applied  to  insulation  measurement,  see  Chapter  V. 

230.  Ballistic  Galvanometer.  A  current  is  often  sent 
through  a  galvanometer  which  flows  for  an  exceedingly 
short  time  only.  This  is  the  case  with  many  induced 
currents  and  condenser  discharges.  Since  the  current 
in  such  cases  does  not  flow  long  enough  to  be  measured, 
a  galvanometer  has  been  devised,  the  deflections  of  which 
are  proportional  to  the  quantity  of  electricity,  or  the  charge, 
passing  through  the  coil.  These  instruments  are  said  to 
"  throw  "  rather  than  "  deflect/'  and  for  this  reason  are 
called  BALLISTIC  GALVANOMETERS.  They  are  made  with 
a  heavier  moving  coil  and  can  be  damped  magnetically, 
without  destroying  the  ratio  between  the  "  charge  "  and 
"  throw."  Such  an  instrument  is  used  in  capacity  meas- 
urements as  follows. 

A  condenser  of  known  capacity  is  discharged  through 
the  ballistic  galvanometer  and  the  "  throw  "  noted.  Then 
a  source  of  unknown  capacity,  charged  to  same  voltage,  is 
discharged  through  the  galvanometer  and  the  throw  noted. 
The  capacity  of  the  two  sources  are  to  each  other  as  their 
respective  galvanometer  throws.  See  Chapter  XL 

231.  Thermo-Electric  Effects.  Before  going  further  into 
the  structural  details  of  electrical  measuring  instruments, 
it  is  necessary  to  consider  briefly  some  phenomena  which 
may  affect  the  choice  of  materials  employed. 

It  has  been  discovered  that  if  two  metals  arc  employed 
in  constructing  an  electric  circuit  and  one  juncture  of  the 
metals  is  heated  to  a  higher  temperature  than  the  other 
juncture,  that  an  electric  pressure  is  set  up  tending  to 
cause  a  current  to  flow.  The  metals  may  be  welded, 
soldered,  or  merely  held  together  by  mechanical  pressure; 
the  E.M.F.  depends  entirely  upon  the  materials  selected 


398  ELEMENTS  OF   ELECTRICITY 

and  the  difference  in  temperature  of  the  two  junctures, 
varying  greatly  with  the  different  combinations  of  materials 
and  being  almost  directly  proportional  to  the  difference 
in  temperature  of  the  joints.  Consider  the  electric  circuit 
in  Fig.  277;  BA  is  a  bismuth  bar,  joined  at  A  to  AC,  a  bar 
of  antimony;  two  metals  often  chosen  because  of  the  high 
value  of  the  thermal  E.M.F.  per  degree  difference  of  tem- 
perature of  junctures.  When  heated  at  point  A,  a  current 
will  flow  in  the  direction  BAC,  because  that  juncture  is 
at  a  higher  temperature  than  the  other  ends  B  and  C. 
If,  however,  the  points  B  and  C  (really  the  other  juncture) 
were  heated,  the  current  "would  flow  in  the  opposite  direction. 
Such  an  arrangement  is  called  a  THERMO-COUPLE.  The 


Sb. 

la 


"A 
Fro.  277. — A  thermo-couple.  FIG.  278. — A  thermo-pile. 

current  will  continue  to  flow  as  long  as  one  juncture  is 
maintained  at  a  higher  temper'ature  than  the  other. 

232.  Thermo-Bolometer.  Pyrometer.  No  successful 
commercial  use  has  yet  been  made  of  this  thermal  E.M.F. 
for  supplying  electric  power,  though  enough  current  can 
be  generated  by  means  of  several  joints  in  series,  alternate 
ones  of  which  are  heated  by  a  gas  flame,  to  supply  electro- 
plating vats.  The  materials  used  in  such  a  device  in  a 
short  time  seem  to  undergo  a  molecular  change  in  their 
structure  which  greatly  diminishes  their  thermo-electric 
power. 

The  thermo-electric  E.M.F.  of  a  pair  of  bismuth  and 
antimony  bars  is  only  about  .1  millivolt  for  every  degree 
centigrade  that  one  juncture  is  above  the  other.  Thermo- 
piles therefore  have  to  be  built  up  of  many  such  pairs  m 
series,  as  in  Fig.  278,  and  one  set  of  junctures  must  be 


ELECTRICAL  MEASURING  INSTRUMENTS 


399 


maintained  at  a  temperature-  considerably  above  the  other 
set,  in  order  to  procure  a  usable  amount  of  pressure. 

Use  is  made  of  this  thermal  E.M.F.  in  making  instruments 
to  measure  very  minute  differences  in  temperature,  and 
also  very  high  or  very  low  temperatures. 

A  thermo-pile  such  as  that  in  Fig.  278  may  be  used  as 
a  sensitive  instrument  for  detecting  small  differences  of 
temperature.  If  the  temperature  of  the  joints  A  is  but 
a  small  fraction  of  a  degree  above  or  below  that  of  the 


FIG.  279. — The  Bristol  pyrometer  for  measuring  high  temperatures. 

joints  B,  an  E.M.F.  will  be  set  up  which  can  be  detected 
by  the  deflections  of  a  sensitive  galvanometer,  G\.  Since 
the  E.M.F.  set  up  is  nearly  proportional  to  the  difference  in 
temperature  of  the  joints,  the  deflections  of  the  galvanom- 
eter can  be  made  to  measure  this  difference  in  temperature. 

By  inserting  in  the  moving  coil  of  a  sensitive  D' Arson val 
galvanometer,  a  bismuth-antimony  thermo-couple,  as  small 
a  rise  or  fall  in  temperature  as  one-hundred-millionth  of 
a  degree  C.  can  be  measured.  Such  an  instrument  is  called 
a  BOLOMETER. 

For  measuring  high  temperatures  a  thermo-couple  of 
platinum  and  rhodium  is  made  in  such  a  form  that  one 


400 


ELEMENTS  OF  ELECTRICITY 


end,  A,  Fig.  279,  may  be  inserted  into  furnaces  of  molten 
metals,  etc.,  and  be  raised  to  the  temperature  of  the  furnace 
or  metal,  while  the  other  end  remains  at  room  temperature. 
The  thermal  E.M.F.  thus  set  up  causes  a  deflection  in  a 
millivoltmeter  calibrated  to  read  temperature  of  end  A 
of  thermo-couple. 

Figs.  279  and  280  are  representations  of  the  Bristol 
Pyrometer,  which  employs  special  alloys  for  the  thermo- 
couple. Note  also  the  device  for  compensating  for  any 


^URNACEv,    •  J  PATENTED 

SEPARABLE 
JUNCTION 


COLD  END  OF 
THERMO-ELECTRIC  COUPLE 


LEADS  TO  INDICATING  INSTRUMENT 


PATENTED  COMPENSATOR  • 

FIG.  280. — The  electrical  connections  of  a  Bristol  pyrometer. 

change  in  temperature  which  may  take  place  at  cold  end 
of  couple. 

233.  Peltier  and  Thomson  Effects.  It  has  also  been 
discovered  that  the  thermo-electric  effect  is  reversible. 
That  is,  if  we  send  a  current  through  a  thermo-couple, 
it  will  either  heat  or  cool  the  joint,  according  to  the  direction 
of  the  current.  This  is  called  the  PELTIER  effect  and  must 
be  carefully  distinguished  from  the  PR  effect,  which  never 
cools  a  wire,  nor  is  it  reversible. 

Sir  William  Thomson  (Lord  Kelvin)  discovered  that  in 
the  case  of  most  conductors  composed  of  pure  metals,  if 


ELECTRICAL  MEASURING  INSTRUMENTS         401 

one  part  were  raised  to  a  higher  temperature  than  another 
part,  a  thermal  E.M.F.  would  be  set  up  between  these  two 
points.  This  effect  is  also  reversible. 

In  consequence  of  these  discoveries  of  the  thermo-electric 
effects  of  different  metals  and  combinations  of  metals, 
it  is  seen  that  too  great  care  cannot  be  exercised  in  the 
selection  of  proper  materials  to  go  into  the  construction 
of  accurate  electrical  measuring  instruments.  It  would 
not  do  to  have  within  the  instrument  itself  a  source  of 
E.M.F.  which  would  affect  its  indications,  if  the  tempera- 
ture of  parts  of  it  were  changed  from  any  cause. 

AMMETERS 

234.  Resistance    of    Ammeters.     Because  ammeters  are 
placed  in  the  line  and  the  whole  current  must  be  forced 
through  them,  the  resistance   of  these   instruments   must 
be  as  low  as  possible.     For  instance,   the  resistance   of  a 
Weston  D.C.  ammeter  of  10  amperes  capacity  is  less  than 
.005  ohm. 

235.  Type    of   Ammeters.     Solenoidal.     The    simple    ar- 
rangement shown  in  Fig.  281  is  often  used  as  an  ammeter 
when    a    cheap    instrument    for 

rough   measurement   is   desired. 

The     current     flowing     through 

the     low     resistance     coil,     C, 

sucks     the     soft     iron    plunger 

A,  pivoted  at    P,  up    into    the 

coil.      This    causes   the    pointer      FlG.  28L_Ammeter 

to    move    over    a    scale    which 

is  calibrated  by  sending  known  currents  through  coil  C. 

The  deflections  are  nearly  proportional  to  the  square  of  the 

current,  since  the  deflecting  force  depends  upon  the  product 

of  the  strength  of  the  field  times  the  strength  of  the  induced 

magnetism,  both  of  which  are  nearly  proportional  to  the 

current  in   the   coils.      The    control    is  effected   by  means 

of   the    weight   W,  called  a  "  gravity  control  ";  damping 


402 


ELEMENTS  OF  ELECTRICITY 


by  means  of  the  eddy  currents  in  the  plunger  A.  Neither 
the  damping  nor  the  control  is  good,  due  to  the  compara- 
tively large  mass  of  the  moving  parts,  the  large  moment 
of  inertia,  and  the  friction  on  the  pivot. 

This  type  can  be  used  equally  well  on  A.C.  circuits,  if 
calibrated  for  that  purpose.  By  making  the  coil  of  many 
turns  of  fine  wire  the  same  design  can  be  used  as  a  volt- 
meter. 

This  instrument,  with  some  refinements,  is  often  used 
on  switchboard  for  the  measurement  of  heavy  currents. 

Another  form  of  the.  solenoidal  type  is  shown  in  Figs. 
282  and  283. 


FIG.  282. — Ammeter.     Another 
form  of  solenoidal  type. 


FIG.  283. — Side  view  of  ammeter 
in  Fig.  282. 


A  and  B  are  soft  iron  rods.  A  is  stationary,  B  is  pivoted 
at  P.  When  a  current  is  sent  through  the  coil,  the  rods 
become  magnetized,  so  that  like  ends  are  nearest  to  one 
another,  and  thus  B  is  repelled  by  A.  This  causes  a  pointer 
to  move  over  a  scale  calibrated  as  in  above  plunger  form 
of  the  type.  It  also  has  a  gravity  control  and  possesses 
but  slight  advantage  over  the  model  first  described.  It 
can  be  used  on  A.C.  circuit,  and  changed  into  a  voltmeter 
by  changing  coil  C.  The  Western  Electrical  Instrument  Co. 
are  now  manufacturing  A.C.  voltmeters  and  ammeters  in 
which  the  above  principle  is  made  use  of,  though  in  a 
greatly  modified  form.  Fig.  284  shows  one  of  these  instru- 
ments. The  magnetized  iron  parts  corresponding  to  the 


ELECTRICAL  MEASURING  INSTRUMENTS 


403 


rods  A  and  B,  Figs.  282  and  283,  are  not  in  the  form  of 
rods,  and  springs  take  the  place  of  gravity  control. 

236.  Thomson  Inclined  Coil  Ammeter.  The  moving 
element  in  this  meter,  shown  in  Fig.  285.  is  a  light  iron 
vane  set  at  an  angle  to  the  shaft 
to  which  it  is  attached.  The 
shaft  itself  is  set  at  an  angle  to 
the  axis  of  the  stationary  coil. 
When  a  current  is  sent  through 
the  coil,  the  vane  tends  to  swing 
into  such  a  position  that  the  lines 
of  force  passing  through  it  shall 
be  parallel  to  the  lines  within 
the  coil.  This  causes  the  needle 
to  deflect  across  a  scale  calibrated 
as  in  previous  models.  The 

deflecting  force  is  nearly  proportional  to  the  current  squared. 
The  control  is  by  means  of  springs  and  is  very  delicate. 
Damping  is  effected  by  air  vanes.  This  instrument  is  one 
of  the  most  accurate  of  those  of  the  solenoidal  type  and 
can  be  used  on  either  D.C.  or  A.C.  circuits. 


FIG.  284.— Weston  portable  A.  C. 
voltmeter. 


FIG.  285. — Thomson  inclined  coil  meter. 

237.  Hot-Wire  Ammeters.  When  an  electric  current 
passes  through  a  wire,  heat  is  generated.  If  this  heat 
is  allowed  to  raise  the  temperature  of  the  wire,  expansion 
will  take  place.  Hot-wire  ammeters  make  use  of  this 
expansion  of  wire  by  the  heat  generated,  to  measure  the 
electric  current  which  generates  the  heat. 


404 


ELEMENTS  OF  ELECTRICITY 


When  a  current  is  sent  through  wire  A B,  Fig.  286,  the 
heat  causes  it  to  sag.  This  sag  is  taken  up  in  the  arrange- 
ment CF.  The  part  CD  is  a  wire,  DE  a  silk  thread  wound 
around  the  pulley  P;  EF  is  a  spring.  As  the  thread  moves 
around  the  pulley  it  causes  it  to  turn,  and  the  pointer 
attached  moves  over  the  scale.  The  deflections  here  are 
almost  proportional  to  the  square  of  the  current,  since  the 
heating  effect  is  proportional  to  the  current  squared,  so 
the  scale  must  be  calibrated  as  in  solenoidal  instruments. 

The  control  is  effected  by 
means  'of  the  elasticity  of 
the  wire  AB  and  spring  S. 
The  instrument,  however, 
requires  frequent  resetting 
to  zero,  is  very  slow  in 
action  and  unless  special 
attachments  are  used,  b 
affected  by  changes  in 
temperature  of  the  room. 
It  measures  both  A.C.  and 
D.C."  currents,  and  can  be 
used  with  shunts  to  increase  its  range.  It  can  also  be  used 
as  a  voltmeter  by  placing  a  high  resistance  in  series  with  it. 
Instruments  of  this  type  have  come  into  use  again  in  the 
measurement  of  alternating  currents  of  the  high  frequen- 
cies used  in  wireless  telegraphy. 

238.  Two-Coil  Ammeters.  Electro  Dynamometers.  *Fig. 
287  shows  the  general  appearance  of  Siemens'  Electro- 
dynamometer.  Fig.  288  shows  the  arrangement  of  the 
parts  of  the  instrument. 

The  current,  after  flowing  through  the  stationary  coil 
C,  enters  the  moving  coil  D,  by  means  of  the  mercury  cups 
M.  The  action  of  the  magnetic  field  of  the  stationary  coil 
on  the  current  in  the  moving  coil  tends  to  rotate  the  moving 
coil  against  the  spring  R.  The  pointer  P  thus  moves  against 
one  of  the  stops  S.  If  now  the  indicator  7  be  moved  in  the 


FIG.  286. — Hot-wire  meter. 


ELECTRICAL  MEASURING  INSTRUMENTS 


405 


opposite  direction  until  the  pointer  P  floats  again,  then 
the  deflecting  force  is  measured  by  the  amount  the  spring 
has  been  deflected,  in  order  'to  restore  the  moving  coil  to 
its  zero  position.  Tne  indicator  /  shows  this  amount 
on  the  graduated  disk  T.  The  deflecting  force  is  proportional 


Fio.  287. — Siemens'  dynamometer. 


FIG.  288. — Construction  of 
dynamometer. 


to  the  product  of  the.  current  in  the  moving  coil  times  the 
current  in  the  stationary  coil.  Since  these  two  coils  are 
in  series,  the  same  current  flows  through  each,  and  the 
deflecting  force  is  therefore  proportional  to  the  square 
of  the  current.  The  scale  divisions  are,  therefore,  not 
uniform  for  a  direct  reading  instrument.  This  instrument 
can  be  used  on  A.C.  circuits;  since  the  current  reverses  in 
both  coils  at  the  same  time,  and  the  relative  direction 
always  remains  the  same.  Its  indications  of  A^C.  values 
are  correct,  even  though  the  instrument  has  been  calibrated 
on  direct  current.  It  can  be  made  extremely  sensitive 


406  ELEMENTS  OF  ELECTU1CITY 

and  accurate.  It  requires  considerable  skill,  however, 
to  manipulate  it;  accordingly  its  use  is  restricted  to  labora- 
tories. By  making  the  coils  of  sufficiently  high  resistance, 
this  type  may  also  be  used  as  a  voltmeter  with  even  better 
results  than  as  an  ammeter. 

Fig.  289  shows   such    an  arrangement    used  as  a  watt- 
meter.    Note  that  the  scale  divisions  are  uniform. 


FIG.  289. — Wattmeter  constructed  on  dynamometer  principle. 

239.  Permanent  Magnet.  Weston  Type.  The  amme- 
ters most  used  in  this  country  for  measuring  direct  currents 
are  of  the  permanent  magnet  and  moving  coil  type.  The 
Weston  ammeters  are  excellent  examples  of  this  form.  They 
are  merely  a  portable  form  of  a  D'Arsonval  galvanometer. 
The  moving  coil,  however,  instead  of  being 'supported  by 
a  delicate  suspension  wire,  is  as  accurately  fitted  into 
jeweled  bearings  as  the  flywheel  of  a  watch.  The  control 
is  effected  by  means  of  two  springs  which  act  against  each 
other.  This  is  much  more  certain  than  the  torsion  of  a 


ELECTRICAL  MEASURING  INSTRUMENTS         407 

long  suspension  wire  and  less  liable  to  injury.     For  view  of 
moving  element  see  Figs.  29,  290  and  291. 


6 


FIG.  290. — Construction  of  Weston  B.C.  meters.     Permanent  magnet  type. 

As  the  fundamental  principles  of  this  instrument  have 
been  explained  in  Chapter  II,  only  the  finer  points  of  con- 
struction will  be  taken  up  here.  Unlike  the  ammeters 


FIG.  291. — Cut-away  view  of  moving  parts  of  Weston  meter. 

previously  described,  the  scale  divisions  of  the  Weston 
ammeters  are  uniform  throughout.  This  is  brought  about 
by  producing  an  absolutely  uniform  field,  in  which  the 


408 


ELEMENTS  OF  ELECTRICITY 


FIG.  292. — Showing  uniform  radi- 
al field  of  permanent  magnet 
type. 


coil  swings.  A  soft  iron  core  is  placed  between  the 
poles  N  and  S  of  the  permanent  magnet.  This  concen- 
trates and  bends  the  magnetic  lines  in  such  a  way  as  to 
make  them  radial  in  direction  (at  right  angles  to  the  arc 
in  which  the  coil  moves),  as  in  Fig.  292.  In  whatever 

position  the  coil  may  be,  there 
are  always  acting  upon  it  the 
same  number  of  lines  at  right 
angles  to  its  motion.  The  deflec- 
tions are  therefore  exactly  pro- 
portional to  the  current  flowing 
in  the  coil.  The  instrument  is 
made  dead  beat  by  the  damping 
effect  of  eddy  currents  set  up 
during  any  motion  of  the  coil 
in  the  aluminum  bobbin  on  which  the  moving  coil  is 
wound.  See  page  202,  Chapter  VIII. 

We  have  seen  that  these  instruments  are  merely  D'Arson- 
val  galvanometers;  they  could  therefore  be  either  ammeters 
or  voltmeters  according  as  the  resistance  of  the  moving 
coil  is  low  or  high.  As  a  matter  of  fact,  they  are  really  all 
MILLIVOLTMETERS.  That  is,  a  few  thousandths  of  a  volt 
are  sufficient  to  cause  a  maximum  deflection  of  the  needle. 
In  fact,  45  millivolts  or  .045  volt  across  the  moving  coil 
gives  a  full  scale  reading  on  the  more  common  type.  Further- 
more, the  wire  composing  the  moving  coil  is  so  fine  that  it 
can  carry  but  a  very  small  fraction  of  an  ampere,  never 
more  than  .05 .  ampere. 

In  order,  therefore,  to  make  -an  ammeter  out  of  this  device, 
which  shall  carry  and  indicate  a  moderately  large  current, 
a  shunt  must  be  placed  across  the  coil  as  in  Fig.  203.  The 
instrument  now  becomes  a  shunted  galvanometer,  and 
a  small  fraction  only  of  the  main  current  flows  through  the 
moving  coil.  Thus,  according  to  Ohm's  Law,  if  the  shunt 
had  .01  ohm  resistance  and  the  coil  gave  its  maximum 
deflection,  it  would  mean  that  there  was  .045  volt  across 


ELECTRICAL  MEASURING  INSTRUMENTS         409 


a  .01   ohm   resistance  and  that 


.045 


:  4.5  amperes  must  be 


flowing    through    t)ie    shunt.     Of    course    the    instrument 


TIG.  293.— Ammeter  with  shunt,  SSS,  inclosed. 

is  calibrated  so  that  the  scale  reads  not  millivolts,  but  the 
amount  of  current  flowing  through  the  instrument,  both 
moving  coil  and  shunt. 


FIG.  294. — Separate  shunts,  for  ammeters. 

This  shunt  must  be  of  such  a  material  or  combination 
of  materials  that  it  does  not  appreciably  change  in  resistance 
as  its  temperature  changes.  In  other  words,  it  must  have 
a  low  Temperature  Coefficient  of  Resistance.  It  also  must 


410 


ELEMENTS  OF  ELECTRICITY 


have  low  Thermoelectric  power  where  soldered  to  copper. 
Such  a  substance  is  the  alloy  MANGANIN,  invented  by  Dr. 
Edward  Weston.  It  consists  of  copper,  nickel  and  ferro- 
manganese.  Several  alloys  have  nearly  as  low  a  temper- 
ature coefficient,  but  do  not  combine  with  this  a  low  thermo- 
elejctric  power.  The  result  on  the  deflections  of  a  milli- 
voltmeter  of  using  a  metal  having  a  high  thermoelectric 
power  can  readily  be  imagined.  The  invention  of  this 
alloy  for  use  in  shunts,  etc.,  is  one  of  the  causes  of  the  success 
of  the  Weston  instruments. 

When  currents  larger  than  25  amperes  are  to  be  measured, 

the  shunt  S  is  not  placed 
inside  the  ammeter  case  on 
account  of  the  large  amount 
of  heat  energy  which  must  be 
dissipated.  Separate  shunts 
are  made  as  shown  in  Fig. 
204.  An  instrument  and  con- 
necting leads  are  calibrated 
to  be  used  with  the  shunt  as 
shown  in  Fig.  295,  and  the 
FIG.  295.— Method  of  using  separate  deflections  indicate  directly 

shunt. 

the  amperes  flowing  through 

the  line  in  which  the  shunt  is  placed.  The  shunt  and 
millivoltmeter  together  constitute  an  ammeter. 


Example.  What  resistance  must  a  shunt  S  be  in  order  that  it 
may  be  used  with  a  .0450  volt  millivoltmeter  to  measure  100 
amperes?  Resistance  of  moving  coil  and  connecting  leads  is 
1.000  ohm.  .' 

Maximum  current  through  coil  =  *—     —  =  .045  ampere. 

1.000 

"      shunt  =  100  -  .045  =  99.955  amperes. 
Resistance  of  shunt  must  be,  by  Ohm's  Law, 

045 

--.0004502  ohm. 


99.955 


ELECTRICAL  MEASURING  INSTRUMENTS         411 

Problem  5-14.  What  resistance  must  a  shunt  be,  in  order  that 
it  may  be  used  with  a  millivoltmeter  of  50  millivolts  range,  to 
measure  5  amperes?  Resistance  of  moving  coil  and  leads  =  1 .025 
ohms. 

Problem  6-14.  If  leads  were  used  which  increased  the  "  coil 
and  lead  "  resistance  to  1.125  ohms,  what  error  would  be  intro- 
duced at  full  scale  reading  in  Problem  5-14,  using  same  shunt? 

Problem  7-14.  Assume  a  copper  wire  is  used  for  the  shunt  in 
Problem  5-14,  and  the  temperature  of  it  rises  10°  C.  during  a 
reading.  What  error  is  introduced  at  full  scale  reading? 

Problem  8-14.  What  would  the  true  reading  be  when  instru- 
ment in  Problem  7-14  read  2  amperes? 

240.  Multiplier.  The  millivoltmeter  described  above 
can  be  used  for  the  measurement  of  voltage,  by  placing 
a  high  resistance  in  series  with  the  moving  coil.  This 
series  resistance  is  called  a  MULTIPLER.  Whenever  the 
instrument  is  to  be  used  entirely  as  a  voltmeter  of  given 
range,  the  multiplier  is  placed  inside  the  case,  and  the 
instrument  is  called,  a  voltmeter. 

Suppose  the  millivoltmeter  of  example,  page  410,  having  a  resist- 
ance of  1  ohm,  were  to  be  used  as  a  voltmeter  of  loO  volt  range. 
What  resistance  must  be  placed  in  series  with  the  coil?  Assume 
.0075  ampere  to  pass  through  meter  at  maximum  reading. 

By   Ohm's   law.      Resistance    of  coil   and  series  resistance  = 

7^=20000  ohms. 
.007o 

Resistance  of  series  resistance  =  20, 000  — 1  =  19,999  ohms. 
Weston  voltmeters  of  the  common  type  have  approximately  100 
ohms  to  the  volt  (maximum  scale  reading). 

Problem  9-14.  The  coil  of  a  Weston  millivoltmeter  of  50 
millivolts  capacity  has  5  ohms  resistance.  What  resistance 
multiplier  must.be  placed  in  series  with  it  in  order  that  it  may 
have  a  maximum  capacity  of  150  volts? 

Problem  10-14.  What  resistance  shunt  must  be  used  with  milli- 
voltmeter in  Problem  9-14,  in  order  that  it  may  read  150  amperes? 
Maximum  current  through  coil  should  be  .01  ampere. 

Problem  11-14.  A  voltmeter  has  a  range  of  100  volts. 
Resistance  is  10,000  ohms.  What  resistance  must  a  multiplier 
be,  in  order  that  the  voltmeter  may  have  a  range  of  300  volts? 


412 


ELEMENTS  OF   ELECTRICITY 


SS,  Stops 
for  Pointer.?. 


Problem  12-14.  Assume  multiplier  in  Problem  11-14  to  be  made 
of  pure  copper.  What  error  is  introduced  by  a  change  of  tempera- 
ture of  8°  C.  if  instrument  is  reading  250  volts? 

Problem  13-14.  If  multiplier  in  Problem  11-14  were  made 
of  German  silver,  what  error  is  introduced  by  a  change  of  8°  C., 
when  instrument  reads  250  volts?  Temperature  coefficient  of 
resistance  for  German  silver  is  .00025. 

241.  Electrostatic  Voltmeter.  One  of  the  disadvantages 
common  to  all  types  of  voltmeters  described  above,  is 
that  they  all  take  power  to  operate  them.  Fig.  296  shows 
an  example  of  the  electrostatic  type  of  voltmeter  which 
takes  no  power  from  line.  It  is  based  on  the  princi- 
ple that  unlike  electrostatic 
charges  of  electricity  attract 
each  other.  The  stationary 
vanes  BB  are  connected  to  one 
side  of  the  line  and  the  mov- 
able vanes  A  A  to  the  other 
side.  The  vanes  A  A  are  now 
attracted  to  the  vanes  BB  in 
direct  proportion  to  the  charge 
FIG.  296.— Construction  of  electrostatic  on  them.  Since  the  charge 

voltmeter. 

upon  each  depends  upon  the 

voltage  between  them,  the  deflecting  force  is  approximately 
proportional  to  the  voltage.  Of  course  as  the  vanes  A  A 
approach  nearer  the  vanes  BB  the  attraction  becomes  greater 
because  of  the  decreased  distance  between  them.  The  scale 
divisions  therefore  are  not  uniform.  A  special  static  volt- 
meter has  been  constructed  in  which  the  moving  vanes  float 
in  oil.  In  this  instrument  the  oil  acts  as  a  damper  and  a 
dielectric.  It  can  be  used  on  higher  voltages  because  of  the 
greater  DIELECTRIC  STRENGTH,  i.e.,  insulating  strength,  of  the 
oil,  and  is  more  sensitive  because  of  the  greater  DIELECTRIC 
POWER  of  the  oil.  Electrostatic  voltmeters  of  course  can  be 
used  equally  well  on  D.C.  and  A.C.  circuits.  They  are  more 
often  used  in  laboratories  than  on  commercial  switchboards. 


ELECTRICAL  MEASURING  INSTRUMENTS         413 

242.  Best  Arrangements  of  Voltmeters  and  Ammeters 
to  Measure  Power,  etc.  Since  all  ammeters  and  all  volt- 
meters (except  the  electrostatic)  consume  power  when  in 
use,  and  thus  introduce  some  error,  it  is  well  to  consider 
the  ways  in  which  they  may  be  arranged  to  make  this  error 
as  small  as  possible. 

Suppose  it  is  desired  to  measure  the  voltage  and  amperage 
of  an  ordinary  incandescent  lamp.     Assume  the  connections 
are  made  as  in  Fig.  297.     A  is  a  Weston  milliammeter, 
resistance  .045   ohm.     V  is  a 
Weston  voltmeter,  resistance 
15,000  ohms.      Assume  both 
instruments  to    be    correctly 
calibrated.     Suppose  the  volt- 
meter   reads    110   volts    and 

the      ammeter      .500      ampere.       FIG.  29?.— Ammeter  measures  current 

in  lamp  plus  current  in  voltmeter. 

The    voltmeter    would    read 

the  voltage  across  the  lamp  correctly  because  it  is  directly 
across  the  lamp  L.  But  the  ammeter  measures  the  cur- 
rent flowing  through  both  the  lamp  and  the  voltmeter. 
It  must,  therefore,  read  too  high.  Let  us  see  how  much 
too  high.  The  current  through  the  voltmeter,  by  Ohm's 
law,  equals, 

0.0073  ampere. 

The  current  through  the  lamp  L  equals, 

.500 -.0073  =.493  ampere. 
Per  cent  error, 

.0073     „ 

~T93~ =  *  per        * 

The  error  of  1J  per  cent  is  altogether  too  large  for  any  such 
simple  measurement. 

Suppose  we  connect  up  the  same  instruments  as  in  Fig. 
298,  with  the  voltmeter  around  both  the  lamp  and  ammeter. 
Assume  the  same  readings  as  before. 


414 


ELEMENTS  OF  ELECTRICITY 


The  ammeter  now  reads  the  current  of  the  lamp  L  only, 
and  is  therefore  correct.  But  the  voltmeter  reads  the 
voltage  across  both  the  lamp  and  ammeter  and  therefore 
reads  too  high.  Let  us  see  how  much  too  high. 

Voltage  across  ammeter, 

.500 X. 045  =.0225  volt. 
Voltage  across  lamp, 

1 10  -  .0225  =  109.98  volts. 
Per  cent  error, 

.0225 


109.98 


no  Volts 


=  .0002  or  -5^  of  1  per  cent. 


.5  Amps. 


FIG.   298. — Voltmeter  measures   drop 
across  lamp  plus  drop  across  ammeter. 


FIG.   299. — Voltmeter   measures   drop 
across  ammeter  and  resistance  R. 


This  error  is  allowable  in  any  grade  of  commercial  work 
and  is  too  small  to  be  considered.  It  is  evident  that  when 
measuring  a  low  current  and  high  voltage,  the  voltmeter 
should  be  placed  around  both  the  ammeter  and  the  apparatus 
under  test.  This  is  because  the  voltage  across  the  ammeter 
is  too  small  to  appreciably  affect  the  reading  of  the  high 
reading  voltmeter. 

Suppose,  however,  that  we  wish  to  measure  the  voltage 
and  amperage  in  a  low  resistance  armature  circuit  R. 

Assume  that  they  are  connected  as  in  Fig.  299,  which 
arrangement,  we  found,  had  the  smaller  error  in  the  pre- 
vious case. 

V  is  a  3-volt  Weston  voltmeter  of  300  ohms  resistance; 
A  a  Weston  ammeter  of  .0009  ohm  resistance.  Assume  the 
voltmeter  reads  2.00  volts  and  the  ammeter  50.0  amperes. 


ELECTRICAL  MEASURING   INSTRUMENTS         415 

The  ammeter  will  read  correctly  the  current  through  R, 
but  the  voltmeter  will  read  the  voltage  across  both  R  and 
the  ammeter.  It  will  thus  read  too  high.  This  error 
can  be  computed  as  before. 

Voltage  across  ammeter, 

.0009  X  50.0  =  .0450  volt. 
Voltage  across  R, 

2.00  -  .0450  =  1.955  volts. 
Per  cent  error, 

r3r2-3percent- 

This  error  is  too  high  for  most  purposes. 

Suppose  we  arrange  the  instruments  as  in  Fig.  300, 
and  assume  the  readings  to 
be  as  before.  The  voltmeter 
now  reads  correctly,  but 
the  ammeter  reads  both  the 
current  through  R  and  the 
voltmeter.  The  ammeter  read- 
ing is  then  tOO  high.  The  FIG.  300. — Ammeter  measures  current 

through  resistance  R  and  voltmeter. 

error  is  found  as  before. 
Current  through  voltmeter, 

2  00 

^=. 00667  ampere; 

Current  through  R, 

50.0 -.00667  =49.993  amperes. 
Per  cent  error, 

=  .00013  or  about  T^  of  1  per  cent. 
*A  y.  v/o 

This  error  is  so  small  that  it  is  allowable  in  all  commercial 
work.  It  is  evident  that  when  measuring  the  power  con- 
sumed by  a  piece  of  apparatus,  through  which  a  large  cur- 


416  ELEMENTS  OF  ELECTRICITY 

rent  at  low  voltage  is  flowing,  the  voltmeter  should  be 
placed  immediately  across  the  piece  of  apparatus  under  test, 
and  not  across  the  ammeter  also.  The  reason  for  this  is 
that  too  small  amount  of  current  flows  through  the  volt- 
meter to  appreciably  affect  the  reading  of  the  large  current 
ammeter. 

243.  Summary   Concerning   Voltmeters   and   Ammeters. 

(1)  SOLENOIDAL  INSTRUMENTS  (containing  soft  iron). 

(a)  Are  simple   and  cheap. 

(b)  Can  be   used  on  both  A.C.    and   D.C.   circuits. 

though  must  be  calibrated  for  each  separately. 

(c)  Are  not  particularly  accurate. 

(2)  HOT  WIRE. 

(a)  Are  simple  and  cheap. 

(b)  Can  be  used  interchangeably  on  both  A.C.  and 

D.C.  circuits. 

(c)  Slow  acting  and  "  dead  beat." 

(d)  Uncertain  control,  but  fairly  accurate. 

(3)  ELECTRO-DYNAMOMETERS. 

(a)  Are  simple. 

(b)  Not  easily  manipulated. 

(c)  Accurate. 

(d)  Can  be  used  interchangeably  on  A.C.  and  D.C. 

(4)  PERMANENT  MAGNET  (moving  coil). 

(a)  Extremely  accurate. 

(b)  Dead  beat. 

(c)  Scale  uniform. 

(d)  Can  be  used  on  D.C.  circuits  only. 

(e)  Positive   control. 

(5)  ELECTROSTATIC. 
(a)  Voltmeter  only. 

(6)  Can  be  used  interchangeably  on  A.C.  and  D.C. 

(c)  Not  particularly  accurate. 

(d)  Takes  no  power. 

(e)  Best    instrument     available     for    high     voltage 

measurement. 


ELECTRICAL  MEASURING  INSTRUMENTS          417 

244.  Indicating  Wattmeter.     Electro-Dynamometer.     In 

discussing  the  electro-dynamometer  we  have  seen  that 
the  deflecting  force  is  proportional  to  the  product  of  the 
current  in  the  stationary  coil  times  the  current  in  the 
moving  coil.  When  used  therefore  as  an  ammeter  or  a 
voltmeter  the  current  in  each  coil  would  be  the  same  and 
the  deflections  proportional  to  the  square  of  the  current  in 
either. 

If,   however,    we    wish   to   measure    the    power    (watts) 
consumed  in  a  given  circuit  we  may  send  the  main  current 


FIG.  301. — Weston  indicating  wattmeter     Note  heavy  terminals  for  current  leads. 

through  the  stationary  coil,  which  we  would  make  oi 
low  resistance,  and  have  the  current  in  the  moving  coil, 
which  would  be  of  high  resistance,  proportional  to  the 
voltage.  That  is,  we  would  put  the  stationary  coil  in, 
series  in  the  line  as  an  ammeter  and  the  moving  coil  across 
the  line  as  a  voltmeter.  The  deflections  would  then  be 
proportional  to  the  product  of  current  times  the  voltage, 
or  the  watts,  and  the  instrument  would  be  a  WATTMETER. 
The  Weston  wattmeter,  constructed  on  the  above  plan, 
is  shown  in  Fig.  301.  The  construction  of  the  instrument 
is  like  that  of  their  permanent  magnet  ammeters  and 
voltmeters.  The  main  difference  is,  that  the  stationary 


418  ELEMENTS  OF  ELECTRICITY 

coils  cc,  Fig.  302,  of  this  instrument  take  the  place  of  the 
permanent  magnets  in  the  other. 

Suppose  the  power  consumed  by  the  arc  lamp  L  were 
to  be  measured.  The  stationary  coils  cc  of  low  resistance 
are  placed  in  series  with  the  arc,  and  a  magnetic  field  is 
set  up  proportional  to  the  current  flowing  through  the  arc. 
The  moving  coil,  with  a  high  resistance  in  series,  is  placed 
across  the  arc  like  the  moving  coil  of  a  voltmeter.  The 
current  through  this  coil  is  proportional  to  the  voltage 
across  the  arc.  The  deflections  then  are  proportional  to 
the  product  of  the  current  through  the  arc  by  the  voltage 
across  the  arc,  or  the  watts.  The  scale  is  graduated  to  read 
directly  in  kilowatts. 

245.  Compensation  in  a  Wattmeter.  By  referring  to 
the  diagram  of  Fig.  302,  it  will  be  seen  that  the  wattmeter 
will  read  too  high  when  connected  as  shown,  because  the 
current  flowing  in  the  stationary  coils  is  not  only  that  through 
the  arc,  but  also  the  current  taken  by  the  moving  coil. 
Thus  the  current  is  too  large.  A  correction  can  easily 
be  made  for  this  by  subtracting  from  the  reading  the  power 
-consumed  by  the  moving  coil.  This  value  should  be  found 
as  follows: 

If  Rm  be  the  resistance  of  moving  coil  and  E  the  voltage 
across  the  lamp,  and  therefore  also  across  the  coil,  the 

E2 
power  consumed  by  the  coil  would  be  -g- . 

tf>m 

If,  however,  we  should  connect  the  voltage  coil  of  the 
wattmeter  across  both  the  lamp  and  the  current  coils, 
then  .the  current  flowing  through  the  stationary  coils 
would  be  that  of  the  lamp  only.  But  the  voltage  across 
the  moving  coil  would  be  the  voltage  across  both  the 
lamp  and  the  stationary  coils.  The  wattmeter  would  still 
read  too  high  because  this  voltage  is  higher  than  the 
lamp  voltage.  The  power  measured  would  be  the  power 
consumed  by  the  lamp  plus  the  power  consumed  by  the 
stationary  coils.  The  correction  can  be  made  by  sub- 


ELECTRICAL   MEASURING  INSTRUMENTS 


419 


tracting  from  the  reading  the  power  consumed  by  the 
stationary  coils.  This  correction  is  found  as  follows: 

Let  I  be  current  through  stationary  coils,  and  Rs  be 
resistance  of  stationary  coils.  Then  power  consumed  by 
stationary  coils  =I2RS. 

In  using  the  West  on  wattmeter  neither  of  these  corrections 
has  to  be  made,  because  of  a  special  compensating  device 
shown  in  Fig.  303.  The  voltmeter,  or  moving  coil,  terminals 
arc  connected  as  in  Fig.  302,  but  .a  compensating  coil,  M, 
is  placed  in  series  with  the  moving  coil.  The  field  of  this 
coil  opposes  the  field  of  the  stationary  coils  cc,  and  weakens 


FIG.  302. — Diagram  of  wattmeter 
without  compensating  coil. 


FIG.  303.— Wattmeter  equipped 
with  compensating  coil. 


it  by  an  amount  proportional  to  the  current  in  the  moving 
coil.  The  field  then  is  exactly  proportional  to  the  current 
flowing  through  the  lamp  alone,  and  the  instrument  now 
indicates  the  watts  consumed  by  the  lamp  alone. 

In  some  cases  it  is  desirable  not  to  use  this  compensating 
coil,  but  to  apply  corrections  computed  as  above.  For 
this  reason  Weston  wattmeters  have  an  extra  binding 
post,  marked  "  Ind  "  (for  Independent),  which  cuts  out 
this  coil.  In  almost  all  commercial  tests,  however,  the 
regular  connections  are  used?  which  include  the  compen- 
sating coil. 


420  ELEMENTS  OF  ELECTRICITY 

Of  course  this  wattmeter  can  be  used  on  A.C.  as  well 
as  D.C.  circuits,  one  calibration  doing  for  both. 

246.  Thomson  Integrating  Wattmeter  or  Watthour 
Meter.  It  has  been  stated  in  Chapter  IV  that  electric 
energy  is  paid  for  by  the  kilowatt-hour;  one  kilowatt-hour 
being  the  quantity  of  electric  energy  consumed  when  power 
is  consumed  for  one  hour  at  the  rate  of  one  kilowatt.  An 
interesting  instrument  called  an  integrating  wattmeter 


FIG.  304. — Thomson  watthour  meter. 

(since  it  adds  up  the  work  done  at  all  instants)  is  used  to 
measure  electric  energy. 

An  illustration  of  this  instrument  is  shown  in  Fig.  304. 
It  is  primarily  a  small  shunt  motor,  the  armature  of  which 
revolves  at  a  speed  proportional  to  the  rate  at  which 
electric  energy  is  passing  through  it.  The  armature  is 
geared  to  recording  dials,  arranged  like  the  dials  on  a  gas 
meter,  which  register  the  number  of  kilowatt-hours  of 
energy  which  have  passed  through  the  meter. 


ELECTRICAL  MEASURING  INSTRUMENTS          421 

The  field  of  this  instrument  is  made  by  current  in  the 
stationary  coils  FF,  Fig.  304,  which  are  in  series  with  the 
line.  There  being  no  iron  in  the  field,  the  field  strength  is 
proportional  to  the  current  flowing  in  the  main  line.  The 
armature  A  is  connected,  by  means  of  a  commutator,  b, 
across  the  line,  so  that  the  current  in  the  armature  is  pro- 
portional to  the  voltage  across  the  line. 

The  torque,  then,  must  be  proportional  to  the  product 
of  the  voltage  times  the  current,  or  the  watts  in  the  line. 

The  friction  is  reduced  to  a  minimum  by  means  of  jeweled 
bearings.  Since  there  is  no  iron  in  the  magnetic  circuit, 
the  field  is  weak,  and  therefore  no  counter  E.M.F.  of  any 
appreciable  value  is  set  up  by  the  rotation  of  the  armature. 
Under  these  conditions  the  armature  would  race  if  there 
were  no  retarding  force  applied  to  the  armature. 

This  retarding  force  must  be  proportional  to  the  speed 
so  that  it  will  decrease  as  the  speed  decreases.  Such  a 
force  is  secured  by  attaching  an  aluminum  disk,  Z),  to  the 
armature,  and  causing  it  to  rotate  between  the  poles  of 
permanent  magnets  M,  as  the  armature  rotates.  This 
motion  sets  up  eddy  currents  in  the  aluminum  diek  which 
arc  proportional  to  the  speed  of  the  disk.  The  retarding 
torque  of  these  currents  must  then  be  proportional  to  the 
speed  of  the  disk,  as  explained  in  Chapter  VIII. 

Thus  the  speed  of  the  armature  will  at  all  times  be  pro- 
portional to  the  torque.  To  illustrate  this,  suppose  we 
assume  the  driving  torque  to  be  increased  suddenly;  the 
speed  would  naturally  rise,  but  as  the  speed  rises  the 
opposing  torque  due  to  the  increased  eddy  currents  also 
rises  until  it  is  equal  to  the  driving  torque  and  the  arma- 
ture speed  becomes  constant.  The  driving  torque  and  the 
retarding  torque  are  thus  always  equal. 

The  above  discussion  is  correct  only  when  we  consider 
the  armature  to  rotate  without  any  friction.  Since  there 
is  always  some  friction  present,  some  arrangement  must 
be  made  to  compensate  for  it.  This  is  done  as  follows :  A 


422 


ELEMENTS  OF  ELECTRICITY 


small  coil  C,  Ffg.  304a,  is  put  in  series  with  the  armature 
and  placed  so  as  to  strengthen  the  field  due  to  the  coils 
FF.  The  position  of  this  coil  is  adjustable  so  that  it  can 


Fi«.  304a. — Adjustable  coil  for  compensating  for  friction. 

be  moved  toward,  or  away  from,  the  armature  as  the  need 
may  be.  It  is  placed  at  a  distance  such  that  the  field 
due  to  it  is  not  strong  enough  to  cause  the  armature  to 
rotate  when  no  current  is  flowing  through  the  main  field 
coils  FF,  but  still  the  lightest  current  through  these  coils 

will  be  sufficient  to  start 
the  motor.  This  compen- 
sating coil,  then,  just  over- 
comes the  friction  of  the 
armature. 

If  the  compensating  coil 
is  too  near  the  main  coils, 
and  the  meter  runs  when 
no  current  is  being  used,  it 

FIG.  305.— Watt-hour  meter  connected  to     IS    Said     to     "  Creep."      This 
line  to  measure  power  consumed  by  set  ,-,         v    ••  .    , 

of  lamps.  causes  the  dials  to  register 

more  energy  than  was  used 

by  the  consumer.  Fig.  305  shows  how  a  watthour  meter 
is  connected  to  the  circuit  in  order  to  measure  the  power 
consumed  by  a  set  of  lamps,  4 


ELECTRICAL  MEASURING  INSTRUMENTS         423 

The  question  is  often  asked:  If  this  meter  is  a  shunt  motor, 
why  is  it  that  the  armature  speed  increases  as  the  field  increases, 
when  it  is  a  well-known  fact  that  the  armature  speed  of  a  shunt 
motor  decreases  as  the  field  increases?  The  answer  to  this  question 
is  simple,  when  we  consider  the  reason  why  a  shunt  motor  de- 
creases in  speed  as  the  field  strength  increases.  The  increase 
in  the  field  strength  of  a  shunt  motor  is  not  the  direct  cause  of 
the  decrease  in  speed.  The  large  decrease  in  the  armature  current, 
due  to  the  increased  counter  E.M.F.,  is  the  reason  for  this  falling 
off  in  speed.  In  the  case  of  the  watthour  meter,  there  is  no  counter 
E.M.F.  of  any  appreciable  value  as  compared  with  the  impressed 
E.M.F.  Thus  any  increase  in  the  counter  E.M.F.  does  not  appreci- 
ably decrease  the  armature  current.  An  increase  in  the  field  strength, 
therefore,  accompanied  by  no  decrease  in  the  armature  current, 
causes  a  larger  torque,  and,  therefore,  increased  speed. 

247.  Voltameter.     We  have  seen  that  a  current  of  one 
ampere  flowing-  for  an  hour  deposits  a  given  amount   of 
metal   from   a   given   solution.     Wherever  the   voltage   is 
constant,  use  can  be  made  of  this  fact  to  measure  the 
total  amount  of  energy  consumed  in  a  given  time.     From 
the   amount  of  metal  deposited,  the  number  of  ampere- 
hours  may  be  computed.     The  number  of  watt-hours  is 
then  found  by  multiplying  by  the  known  constant  voltage. 

The  apparatus'  required  consists  of  a  jar  of  electrolyte 
and  two  plates.  The  "  gain  "  or  negative  plate  is  weighed 
before  and  after  the  run.  The  gain  in  weight  divided  by 
the  electrochemical  equivalent  gives  the  number  of  ampere- 
hours.  The  area  of  the  gain  plate  in  a  copper  voltmeter 
should  be  about  4  sq.in.  to  the  ampere. 

248.  Potentiometer.     For  very  precise  measurement  of 
voltage,  the  potentiometer  is  the  instrument  most   used. 
The  principle  on  which  it  operates  may  be  seen  from  an 
inspection  of  Fig.  306. 

Two  sources  of  E.M.F.  cell  I  and  cell  II  are  joined 
in  parallel  and  are  connected  to  the  slide  wire  AB.  Cell  I 
must  be  of  higher  E.M.F.  than  cell  II,  and  be  able  to  deliver 
a  steady  current.  In  series  with  cell  I  is  the  variable 
resistance  R.  Cell  II  is  a  standard  cell,  generally  a  Weston, 
and  must  be  allowed  to  deliver  no  appreciable  current. 


424 


ELEMENTS  OF  ELECTRICITY 


In  series  with  standard  cell  II  is  a  galvanometer  G  and  a 
contact  maker  C.  If  contact  maker  C  is  not  touching  the 
wire  AB,  then  cell  I  delivers  a  current  through  AB  from 
A  to  B.  Cell  II  not  having  a  closed  circuit  delivers  no 
current. 

If  we  place  a  weak  cell  like  cell  II  directly  across  the 
terminals  of  a  strong  cell  like  cell  I,  the  stronger  cell  will 
force  a  current  through  the  weak  cell  in  the  reverse  direction 
to  the  E.M.F.  of  the  weak  cell. 

Suppose  contact  maker  C  touches  the  wire  A  B  at  such  a 
point  that  the  IR  drop  along  AB  is  greater  than  the  E.M.F. 


FIG.  306. — Construction  of  potentiometer  for  measuring  low  voltages. 

of  cell  II.  The  two  cells,  being  in  parallel,  the  stronger 
cell  I  will  force  a  current  in  the  reverse  direction  through 
cell  II.  For  while  cell  II  is  not  directly  across  cell  I,  still 
it  is  across  enough  of  the  voltage  of  cell  I  to  have  its 
weak  E.M.F.  overpowered.  The  galvanometer  G  will 
then  deflect  in  a  certain  direction.  Cell  I  is  then  deliver- 
ing two  parallel  currents,  one  from  A  to  B,  through  the 
slide  wire,  and  the  other  from  A  to  C,  through  cell  II, 
against  the  E.M.F.  of  cell  II. 

On  the  other  hand,  suppose  we  touch  C  to  AB  at  such 
a  point  that  the  IR  drop  from  A  to  C  is  less  than  the  E.M.F. 
of  cell  II.  Then  cell  II  aids  cell  I  in  sending  current  from 
A  to  (7,  and  raises  the  IR  drop  by  increasing  the  current  7. 
Current  will  now  flow  out  of  cell  II,  and  the  galvanometer 
will  be  deflected  in  the  opposite  direction. 


ELECTRICAL  MEASURING  INSTRUMENTS         425 

Suppose,  however,  that  we  touch  C  to  the  wire  AB  at 
such  a  place  that  the  IR  drop  along  AB  will  be  exactly  equal 
to  the  E.M.F.  of  cell  II.  The  force  tending  to  send  a 
current  in  the  reverse  direction  through  cell  II  will  be 
exactly  counterbalanced  by  the  E.M.F.  of  the  cell,  and 
no  current  will  flow  in  either  direction  through  cell  II.  The 
galvanometer  will  then  show  no  deflection  and  the  poten- 
tiometer is  said  to  be  "  balanced." 

As  the  value  of  the  E.M.F.  of  the  cell  II  is  known  accu- 
rately, the  value  of  the  IR  drop  from  A  to  C  is  known 
accurately.  A  cell  of  unknown  E.M.F.  can  now  be  put 
in  the  place  of  cell  II  and  a  balance  obtained  as  before. 
Since  the  IR  drop  along  the  wire  is  proportional  to  its 
length  (I  remaining  constant),  the  E.M.F.  of  the  unknown 
cell  is  to  the  E.M.F.  of  cell  II  as  the  distances  AC,  in  the 
respective  settings.  By  means  of  the  rheostat  R,  the 
IR  drop  in  AC  can  be  regulated  so  that  the  scale  divisions 
along  the  slide  wire  are  some  decimal  multiple  of  the  drop 
along  the  wire.  For  instance,  in  using  a  Weston  cell,  the 
E.M.F.  of  which  is  1.01985  volts,  the  resistance  R  would 
be  regulated  until  the  potentiometer  is  balanced  with 
C  at  1019.9  millimeters  from  A.  Whenever  a  cell  of  un- 
known E.M.F.  was  put  in  the  place  of  cell  II  and  balanced, 
the  distance  of  point  C  from  A  in  millimeters,  divided  by 
1000,  would  be  the  E.M.F.  of  the  unknown  cell.  This  is 
called  "  setting  the  potentiometer." 

NOTE.  The  potentiometer  is  merely  a  special  case  of  two 
generators  in  parallel  feeding  into  a  common  line  AC.  When 
the  voltage  across  the  line  is  less  than  the  E.M.F.  of  each  generator, 
they  both  feed  into  the  line.  When  the  voltage  across  the  line  is 
less  than  the  E.M.F.  of  one  generator  but  greater  than  that  of 
the  other,  the  generator  of  the  higher  E.M.F.  feeds  into  the  line, 
and  backs  up  a  current  through  the  other  generator. 

But  when  the  voltage  across  the  line  happens  to  just  equal  the 
E.M.F.  of  one  generator,  that  generator  neither  delivers  nor  takes 
current.  The  other  generator  feeds  into  the  line. 

The  voltage  of  the  line  could  not  be  greater  than  nor  equal 
to  the  E.M.F.  of  both  generators,  since  one  generator  must  feed 
into  the  line  to  cause  any  drop  across  it. 


426 


ELEMENTS  OF  ELECTRICITY 


249.  Calibration    of    Voltmeter    by    Potentiometer.     In 

order  to  calibrate  a  voltmeter,  the  range  of  which  may 
be  several  times  greater  than  the  greatest  IR  drop  along 
the  slide  wire,  an  arrangement  shown  in  Fig.  306a  is  used. 
The  wire  AB  is  calibrated,  as  described  above,  to  read 
direct,  or  have  some  simple  constant.  The  voltmeter 
is  placed  across  E2;  a  variable  source  of  E.M.F.  For 
readings  which  do  not  exceed  the  IR  drop  along  AB  a 
plug  is  placed  at  D  and  balance  obtained  as  described  above. 
When  the  voltage  desired  for  calibration  becomes  greater 
than  the  IR  drop  along  AB}  the  plug  D  is  withdrawn  and 


20  80  100          800         1000 

Ohms     Ohms   Ohms   Ohms  Ohms 


FIG.  306a. — Potentiometer  for  measuring  high  voltages. 

placed  in  P.  This  gives  an  IR  drop  of  E2  across  the  2000 
ohm  resistances  MN.  If  now  the  plug  F  is  put  in,  but 
half  of  this  voltage  E2  is  put  across  AC.  The  reading  of 
AC,  multiplied  by  2,  would  then  be  the  correct  voltage 
across  E2.  If  the  plug  is  placed  in  K,  for  instance,  only 
Y^-Q-  of  the  voltage  of  E2  is  thrown  across  AC,  and  thus  the 
setting  AC,  multiplied  by  100,  gives  the  true  voltage  E2. 

This  arrangement  enables  voltmeters  to  be  quickly 
and  accurately  calibrated  through  a  wide  range  of  values. 

In  connection  with  a  standard  lesistance,  the  potentiom- 
eter affords  an  accurate  method  for  calibrating  ammeters. 


ELECTRICAL  MEASURING  INSTRUMENTS         427 


SUMMARY  OF  CHAPTER  XIV 

GALVANOMETERS. 

(1)  Tangent, 

(a)  Not  sensitive. 

(2)  Astatic    (Thomson), 

(a)  Sensitive; 

(b)  Not  dead-beat; 

(c)  Slow  action,  and  fair  control. 

(3)  D'Arsonval, 

(a)  Moving  coil,  permanent  magnet ; 

(b)  Sensitive; 

(c)  Dead-beat; 

(d)  Positive    control. 

With  shunts,  galvanometers  can  be  used  to  measure  cur- 
rent; with  series  coil,  to  measure  voltage. 
EQUATION  FOR  SHUNTS. 

Current  through  Galvanometer        S 
Current  through  main  line          S+G* 

SENSIBILITY,  (a)  Number  of  amperes  or  coulombs  to 
produce  a  deflection  of  one  scale  division,  (b)  Number 
of  megohms  which  may  be  placed  in  series  with  galvanometer 
across  one  volt  pressure  and  have  the  instrument  give  a 
deflection  of  one  scale  division. 

A  BALLISTIC  GALVANOMETER.  "Throws"  are  pro- 
portional to  quantity  of  electricity  discharged  through  it. 

THERMO-ELECTRIC  PRESSURE.  Is  set  up  when  one 
juncture  of  two  unlike  metals  is  at  a  higher  temperature  than 
the  other.  Possessed  by  different  combinations  of  metals  in  a 
greatly  varying  degree.  Effect  used  in  pyrometers  to  meas- 
ure high  temperatures.  Such  combinations  of  metals  must 
be  avoided  in  ordinary  ammeters  and  voltmeters. 

AMMETERS  AND  VOLTMETERS.  See  summary  on  page 
416. 

ARRANGEMENT  OF  INSTRUMENTS  FOR  POWER 
MEASUREMENT.  In  using  wattmeters,  or  ammeters  and 
voltmeters,  for  accurate  power  measurement,  they  should 
be  connected  to  the  circuit  in  such  a  way  as  to  cause  as  small 
error  in  readings  as  possible.  In  very  accurate  work  the  errors 
due  to  power  consumed  by  instruments  must  be  corrected. 


428  ELEMENTS   OF  ELECTRICITY 

WATTHOUR  METER— THOMSON.  A  small  shunt  motor, 
the  speed  of  which  is  proportional  to  rate  of  power  con- 
sumption. Armature  current  proportional  to  voltage;  field 
proportional  to  current.  Opposing  torque  supplied  by  eddy- 
currents  in  disk  which  revolves  between  permanent  mag- 
netic poles. 

POTENTIOMETER.  A  device  for  measuring  voltage  or 
E.M.F. ;  takes  no  current  from  source,  the  pressure  of  which 
it  measures.  E.M.F.  to  be  measured  is  balanced  against  the 
fall  of  potential  along  a  wire. 

By  balancing  a  definite  fraction  of  the  E.M.F.  against  this 
fall  of  potential,  high  voltages  may  be  measured,  and  high 
reading  voltmeters  accurately  calibrated. 


PROBLEMS   ON    CHAPTER  XIV 

14-14.  The  scale  of  an  electro-dynamometer  is  divided  into 
360°.  The  instrument  is  to  be  used  as  an  ammeter,  and  2 
amperes  are  found  to  give  a  deflection  of  15°.  How  many  degrees 
deflection  would  the  following  currents  give:  (a)  1  ampere;  (b) 
4  amperes;  (c)  5.5  amperes? 

15-14.  How  many  amperes  would  the  following  deflections 
of  the  electro-dynamometer  of  Problem  14  indicate:  (a)  10°; 
(ft)  40°;  (c)  75°? 

16-14.  It  is  desired  to  measure  the  power  taken  by  an  electric 
device.  Instruments  are  arranged  as  in  Fig.  297.  The  ammeter 
has  a  resistance  of  0.09  ohm;  the  voltmeter  has  a  resistance  of 
16,500  ohms.  Ammeter  reading  =  .465  amperes.  Voltmeter  reads 
112  volts.  If  instruments  are  correct,  what  is  per  cent  error  in 
power  measurement  due  to* the  arrangement  of  the  instruments? 

17-14.  If  above  instruments  were  arranged  as  in  Fig.  298, 
(a)  what  would  each  read?  (b)  What  would  be  the  per  cent 
error  in  power  measurement  for  this  arrangement? 

18-14.  In  measuring  the  core  loss  of  a  5  K.W.  transformer  the 
wattmeter,  ammeter,  and  voltmeter  were  connected  as  shown 
in  Fig.  307.  C  and  D  are  the  ammeter  or  current  connections  to 


ELECTRICAL  MEASURING  INSTRUMENTS 


429 


the  wattmeter,  and  E  and  F  the  voltmeter  or  potential  connections, 
The  compensating  coil  is  not  used. 

Voltmeter  has  a  resistance   of  2200  ohms; 

Current  coil  of  wattmeter,   .0045  ohm; 

Potential   coil  of  wattmeter,   3300  ohms. 
The  readings  were  as  follows: 

Ammeter  =  .634  ampere; 

Voltmeter  =  110  volts ; 

Wattmeter  =  69. 7  watts. 
What  was  the  true  core  loss  in  watts? 


o 

0 
0 

'C  o 

0 

o 

0 

jg 

jO  v 

Jo 

F 

To 
Power 

© 

E 

<^ 

^ 

~*C      D*" 

*—  ^ 

FIG.  307. — Measurement  of  core  loss  with  ammeter,  voltmeter,  and  wattmeter. 

19-14.  If  instruments  used  in  Problem  18  are  connected  to 
measure  transformer  core  loss  as  in  Fig.  307a,  what  will  each 
instrument  indicate? 

20-14.  A  potentiometer  arranged  as  in  Fig.  306a  is  being  used 
to  calibrate  a  voltmeter  V.  1  millimeter  of  the  slide  wire  =  l 
millivolt  drop.  The  slide  C  is  at  1428.3  mm.  when  the  instru- 


E| 

o 


Fi«.  307a. — Another  arrangement  of  ammeter,  voltmeter,  and  wattmeter  for 
measuring  core  loss. 

ment  is  balanced.      If  plug  D  is  in  place,  what  is   the  voltage 
across  J£2?  » 

21-14.  Assume  that  the  same  reading  were  obtained  in  Prob- 
lem 20,  with  plug  D  withdrawn  and  plug  J  in  place.  What  would 
be  voltage  across  EJ  Plug  P  is  in  place. 


CHAPTER    XV 
ALTERNATING    CURRENTS 

Definition:  Cycle;  Frequency;  Phase;  A.C.  Generator;  Vector  Dia- 
grams, Average,  Maximum,  Effective  and  Instantaneous  Values 
of  E.M.F.  and  Current — Computation  of  Current;  Current  and 
Voltage  in  Phase;  Leading  Current;  Lagging  Current — Causes  and 
Effects  of  Lead  and  Lag — Reactance,  Inductance,  and  Capacity — 
Computation  of  Reactance;  Impedance,"  Computation  of  Ohm's 
Law  for  A.C.  Circuits — Series  and  Parallel  Circuits — Power  in  A.C. 
Circuits — General  Law  for  A.C.  Circuits;  when  Current  and 
Voltage  are  in  Phase;  when  out  of  Phase — Power  Factor — Effect 
of  Inductance  and  Capacity  on  Power  Factor — Use  of  Ammeter 
and  Voltmeter  for  A.C.  Power  Measurements — Use  of  Wattmeter. 

250.  Cycle:  Frequency.  An  electric  current  which  flows 
back  and  forth  at  regular  intervals  in  a  circuit  is  called  an 
ALTERNATING  CURRENT.  When  the  current  rises  from 
zero  to  a  maximum,  returns  to  zero,  and  increases  to  a 
maximum  'in  the  opposite  direction  and  finally  returns  to 
zero  again  it  is  said  to  have  completed  a  CYCLE.  For  con- 
venience a  cycle  is  divided  into  360  degrees.  Any  point 
in  the  cycle  is  spoken  of  as  a  certain  PHASE.  Thus  when 
the  cycle  is  half  completed,  it  is  said  to  be  at  the  ISO-degree 
phase;'  when  one-fourth  completed  at  the  90-degree  phase. 
This  cycle  of  changes  is  repeated  over  and  over  again.  Fig. 
331  is  a  graphical  representation  of  the  cycle  of  such  a  cur- 
rent. The  maximum  value  in  one  direction  is  reached  at  the 
90-degree  phase,  and  in  the  other  direction  at  the  270-degree 
phase.  The  number  of  times  this  cycle  takes  place  in  one 
second,  is  called  the  FREQUENCY  of  the  current.  Thus  a 
current  which  rises  to  a  maximum  in  each  direction  60  times 

430 


ALTERNATING   CURRENTS  431 

a  second  is  said  to  make   60  cycles  per  sec.,  or  to  have  a 
frequency  of  60.     The  symbol  for  frequency  is  /. 

251.  Water   Analogy.     If  we  liken  the  flow  of  a  direct 
current  to  the  flow  of  water  in  a  river,  we  may  liken  the 


FIG.  308. — Pipe  circuit  containing  a  pump  with  valves.       Corresponds  to  a  D.C. 
electric  circuit. 

flow  of  an  alternating  current  to  the  ebb  and  flow  of  the 
tide  in  a  narrow  channel.  Of  course  the  frequency  of 
such  a  tidal  flow  would  be  extremely  'small.  Some  ideas 


FIG.  309. — Pipe  circuit  containing  a  valveless  pump.     Corresponds  to  an  A.C. 
electric  circuit. 

of  the  nature  of  direct  currents  and  alternating  currents  can 
be  derived  from  the  study  of  .Figs.  308  and  309. 

Fig.   308  represents   a  pipe   circuit   containing   a  pump 
with  valves  so  arranged  that  the  water  in  the  pipe  flows 


432 


ELEMENTS  OF  ELECTRICITY 


in  one  direction  only,  independent  of  the  direction  of  the 
piston  motion.  This  is  the  water  analogy  to  a  direct 
current  system.  The  pipe  takes  the  place  of  the  wires, 
the  pump  cf  the  generator.  The  valves  of  the  pump 
represent  the  commutator  of  the  generator.  Both  valves  and 
commutator  allow  the  current  to  flow  in  but  one  direction. 
The  pump  in  Fig.  309  has  no  valves.  The  direction 
of  the  current  in  the  pipe  then  depends  upon  the  direction 
of  the  piston  motion.  The  pump  without  valves  represents 
a  generator  without  a  commutator.  Such  a  generator 
would  deliver  an  alternating  current 
to  the  line,  as  explained  in  Chapter 
VII. 

We  will  now  consider,  in  greater 
detail,  the  E.M.F.  and  current  set 
up  in  a  single  coil  armature  of 
such  an  alternating  current  gener- 
ator. 

252.  Alternating  Current  Gene- 
rator. Single  Coil.  Fig.  310  is  a 
simple  diagram  in  perspective  of  a 
single  coil  armature.  An  alternat- 
ing E.M.F.  will  be  set  up  in  the 
coil  if  it  is  revolved  as  explained 
in  Chapter  VII. 

The  following  details  in  connec- 
tion    with     the     E.M.F.     induced 
should  be  studied  carefully. 
When  the  coil  is  in  the  position  shown  in  Fig.  311,  it  is 
cutting  no  lines  of  force.     There  is  therefore  no  voltage 
induced  in  it.     This  position,  or  phase  angle,  is  zero. 

When  the  coil  has  moved  30  degrees  from  the  zero  as 
shown  in  Fig.  312,  it  is  cutting  force  lines,  and  •there  is  an 
E.M.F.  induced  of  OUT  at  A,  and  IN  at  B. 

Suppose  we  plot  the  E.M.F.  as  ordinates  and. the  position 
of  the  coil  in  degrees  from  zero  as  abscissae,  Fig.  316. 


FIG.  310. — Two  pole  generator 
with  single  coil  armature. 


ALTERNATING  CURRENTS 


433 


The  line  e\  represents  the  instantaneous  voltage  at  the 
30°  phase,  or  when  coil  is  at  30°  as  shown  in  Fig.  312.1 


D 


FIG.  311. — Coil  passing  through  zero  po- 
sition and  cutting  no  force  lines.  Fig. 
317  is  vector  diagram  for  E.M.F.  at 
this  instant. 


FIG.  312.— Coil  passing  through  30- 
degree  position.  Fig.  318  is  vector 
diagram  for  E.M.F.  at  this  instant. 


When  the  coil  has  moved  60°  from  the  zero  position  as 
in  Fig.  313,  the  instantaneous  E.M.F.  has  increased  to  a 
value  represented  by  the  line  62,  Fig.  316. 


FIG.  313. — Coil  passing  through  60-degree 
position.  Fig.  319  is  vector  diagram 
for  E.M.F.  at  this  instant. 


FIG.  314. — Coil  passing  through  90- 
degree  position.  Fig.  320  is  vector 
diagram  for  E.M.F.  at  this  instant. 


1  The  angular  position  of  the  coil  and  the  phase  angle  are 
the  same  in  a  two-pole  machine,  but  not  in  others.  When 
the  phase  differs  from  the  angular  position,  it  is  customary 
to  speak  of  the  position  of  the  coil  as  "  so  many  degrees  "  and 
of  the  phase  as  so  many  "  electrical  degrees."  This  causes  no 
confusion  in  actual  practice. 


434 


ELEMENTS  OF  ELECTRICITY 


When  the  coil  has  moved  90°  from  the  zero  position, 
it  is  then  cutting  lines  at  the  fastest  rate  as  it  is  moving 
across  them  at  right  angles,  as  shown  in  Fig.  314.  The 
line  E,  Fig.  316,  now  represents  the  instantaneous  induced 
voltage. 

As  the  coil  moves  on  from  this  point,  the  induced  voltage 
across  it  begins  to  decrease,  and  at  length  becomes  zero 
again,  as  the  coil  reaches  a  point  180°  from  zero  position. 
Lines  e3  and  e4,  Fig.  316,  show  the  instantaneous  value 
of  the  E.M.F.  at  120°  and  150°  respectively.  At  180° 
the  instantaneous  E.M.F.  is  zero,  but 
as  soon  as  it  passes  this  point  the  coil 
begins  to  cut  lines  in  the  opposite 
direction,  and  an  E.M.F.  is  now  in- 
duced which  is  IN  at  A,  and  OUT  at 
B,  as  shown  by  Fig.  315.  The  instan- 
taneous value  of  this  induced  E.M.F. 
is  represented  by  the  line  —  e\  in  Fig. 
316.  The  value  continues  to  increase 


FIG.  3i5.  — Coil   passing     in  this  reverse*  direction,  until  a  posi- 

through  210-degree  posi- 
tion. Fig.  312  is  vector 
diagram  for  E.M.F.  at 
this  instant. 


tion  270°  from  zero  is  reached. 
Here  the  voltage  becomes  a  maxi- 
mum and  is  represented  by  the 

line  —  E,  Fig.  316.  From  here  on,  it  gradually  decreases 
until  it  becomes  zero  again  as  it  reaches  360°  (which  is 
really  the  zero  point  at  which  we  started  to  follow  the 
movement  of  the  coil).  The  lines  -e3  and  —e4  represent 
the  values  of  the  E.M.F.  at  300°  and  330°  respectively. 
The  curve,  Fig.  316,  plotted  from  these  values,  represents 
the  various  instantaneous  values  of  the  E.M.F.  at  all  times, 
during  one  complete  cycle.  Note  that  the  E.M.F.  is  con- 
tinually changing,  and  if  we  wish  to  indicate  its  value  we 
must  state  it,  as  for  a  given  instant  only.  At  any  other 
instant  it  will  be  greater  or  less  than  this  value. 

As   the   armature   turns   around   it   merely   repeats   this 
cycle  of  values,  shown  by  the  vertical  lines  in  Fig.  316. 


AL  TERN  A  TING  C  URREN  TS 


435 


Such  a  curve  is  known  as  a  SINE  CURVE  and  approximates 
more  or  less  closely  the  curves  of  the  E.M.F.  values  of 
most  A.C.  generators. 

253.  Vector  Diagram.  Figs.  317-326  show  another  and 
a  simpler  way  of  representing  the  relation  of  the  E.M.F. 
induced  in  an  armature  coil  to  the  phase  angle,  when  the 
E.M.F.  follows  a  sine  curve. 

The  greatest  value  which  the  E.M.F.  attains  is  repre- 
sented by  the  line  E,  called  a  VECTOR.  This  line  is  supposed 
to  be  rotating  in  a  counter-clockwise  direction  about  the 


30°   60°   90°  120°  150°  l8Q°  210C 
-e\ 


\ 


240°    270°    300° 


-E 


FIG.  316. — Sine  curve  showing  relation  between  E.M.F.  and  phase  (or,  in  this  case, 
angular  position  of  coil). 

end  o.  By  drawing  this  vector  to  some  scale,  and  at  angle 
to  an  axis,  equal  to  phase  angle,  we  determine  the  instan- 
taneous value  at  that  position,  or  phase,  by  measuring 
the  vertical  component  to  the  vector  E.  The  X  axis  has 
been  used  for  convenience  in  Figs.  317-326  to  represent  the 
zero  position  of  the  vector. 

Thus  in  Fig.  317,  which  is  the  vector  diagram  for  coil 
in  position  shown  in  Fig.  311,  the  vector  E  has  not  moved 
from  the  zero  position  and  has  no  vertical  component. 
Thus  the  value  of  the  E.M.F.  in  this  position  is  0.  This 
is  the  value  of  the  E.M.F.,  as  we  have  seen,  when  the  coil 
is  in  the  position  of  Fig.  311. 

In  Fig.  318,  the  vector  diagram  represents  the  instan- 
taneous E.M.F.,  when  the  position  of  the  coil  is  as  shown 


436 


ELEMENTS  OF  ELECTRICITY 


in  Fig.  312.  The  coil  has  moved  30°  from  the  zero  position. 
The  vector  E  is  therefore  at  an  angle  of  30°  to  the  horizontal. 
The  vertical  component  of  E  is  now  e\,  which  equals  the 
instantaneous  value  of  the  induced  E.M.F.,  when  the  coil 


H 


,\ 


FIG.  317. — Vector  diagram  for  E.M.F. 
when  coil  is  passing  through  zero 
position.  See  Fig.  311. 


FIG.  318. — Vector  diagram  when  coil  is 
sing  through  30-degree  position. 
Fig.  312. 


is  30°  from  the   zero   position.     The  vertical   component 
e2  at  60°  is  shown  in  Fig.  319. 

When  the  vector  reaches  the  90°  position,  the  vertical 
component  is  equal  to  the  vector  itself,  and  reaches  its 
maximum  value  when  the  coil  is  90°  from  the  zero  position. 


FIG.  319.— Vector  diagram  for  E.M.F. 
at  60-degree  phase.     See  Fig.  313. 


FIG.  320. — Vector  diagram  for  90-degree 
phase.     See  Fig.  314. 


Fig.  314  represents  the  coil  in  this  position  and  Fig.  320 
represents  the  vector  diagram  of  the  E.M.F.  at  this  time. 
Figs.  322  and  323  are  the  vector  diagrams  for  the  coil 
in  positions  150°  and  180°  respectively.  Note  that  e4  is 
the  instantaneous  value  for  the  E.M.F.  when  the  coil  is 
at  the  150°  point  and  its  value  corresponds  to  E4  of  Fig. 
316,  At  180°  the  instantaneous  value  of  the  E.M.F.  is 


ALTERNATING  CURRENTS 


437 


zero.  Accordingly,  in  the  vector  diagram,  Fig.  323,  there 
is  no  vertical  component  of  E,  showing  that  the  instan- 
taneous value  of  the  E.M.F.  is  zero. 

When  the  coil  passes  the  180°  point,  the  E.M.F.  reverses. 
This  is  shown  in  Fig.  324,  by  the  fact  that  the  line  e\,  which 


Fi«.  321.— Vector  diagram  for  E.M.F. 
at  120-degree  phase. 


FIG.  322. — Vector  diagram   for  E.M.F. 
at  ISO-degree  phase. 


is  the  vertical  component  of  E,  is  below  the  -X"  axis.  At  270°, 
the  instantaneous  value  of  the  E.M.F.  becomes  a  maximum 
in  this  reversed  direction.  This  appears  in  Fig.  325,  where 


FIG.  323. — Vector  diagram  for  E.M.F. 
at  180-degree  phase. 


FIG.  324.— Vector  diagram  for  E.M.F. 
at  210-degree  phase.     See  Fig.  315. 


the  vertical  component  of  E  equals  —  E,  and  is  below  line, 

thus  being  a  minus  quantity. 

In  the  same  manner,  —  e4,  in  Fig.  326,  corresponds  to 
—  64  in  Fig.  316,  and  represents  the  instantaneous  value 
of  the  E.M.F.,  when  the  coil  is  at  the  330°  point. 


438 


ELEMENTS  OF  ELECTRICITY 


Note  carefully: 

(a)  That  the  lines  E,  ei}  e2,  e3,  e±,  and  —  ei,  —  e2,  —  e3, 
—  e4,  represent  merely  instantaneous  values  of  the  E.M.F. 
at  the  various  positions  of  the  coil,  or  at  various  phases  in 
a  cycle. 

(b)  That  E  represents  the  maximum  value  that  these 
instantaneous  values  ever  reach. 

(c)  That   the   maximum  is   reached  twice   during  each 
cycle. 

It  follows  from  this  that  e\,  e2,  and  —e\  —e2,  etc.,  can  be 
regarded  as  merely  certain  fractions  of  the  maximum  value 


FIG.  325.— Vector  diagram   for  E.M.F. 
at  270-degiee  pahse. 


FIG.  326. — Vector  diagram  for  E.M.F. 
at  330-degree  phase. 


E.  Now  we  have  stated  that  the  curve,  Fig.  316,  was 
called  a  sine  curve.  It  is  so  called  because  the  ratio  between 
the  instantaneous  values  e\,  e2,  etc.,  and  the  maximum  value 
E,  equals  the  sine  of  the  phase  angle  at  that  instant. 

Thus  in  Fig.  318,  e\,  the  instantaneous  value  at  the  30° 
phase,  is  a  certain  fraction  of  E.  The  value  of  this  fraction 
is  the  sine  of  the  angle  30°,  since  the  phase  is  30°.  By 
consulting  a  table  of  sines,  we  find  that  the  sin  30°  =  .5. 
Thus  e\  is  i  of  E.  So  if  #  =  150  volts,  the  instantaneous 
value  of  the  voltage  when  only  30°  of  the  cycle  had  been 
completed  would  be  J  of  150  or  75  volts. 

In  Fig.  319,  e2  is  that  fraction  of  E,  which  is  equal  to 
the  sine  of  the  angle  60°,  Sin  60°  =  .866.  Therefore, 


ALTERNATING  CURRENTS  439 

If  7?  =  150  volts, 

c2=. 866X150  =  130  volts. 

254.  Equation   for  Instantaneous  Value  of  E.M.F.    The 

rule  then  for  finding  any  instantaneous  value  of  the  voltage 
in  an  alternating  current  circuit,  when  the  voltage  follows 
the  sine  wave,  is:1 

Multiply  the  maximum  value  of  the  voltage  by  the  sine 
of  the  angle  of  phase.  This  is  generally  stated  in  the  form 
of  an  equation: 

e=E  sin  (/>; 

where      e=  instantaneous  voltage; 
E  =  maximum  voltage ; 
0=  phase  angle  (in  degrees). 


FIG.  327.— Sine  curve  of  E.M.F. 

There  are,  then,  three  general  ways  of   representing   an 
alternating  E.M.F.  and  of  determin- 
ing its  value  at  various  instants. 

(1)  By   the   graphical    representa- 
tion  of  the  SINE  CURVE,  as  in   Fig. 

327.  

(2)  By  the   vector  diagram,  as   in 
Fig.  328. 

(3)  By  an  equation,  as 


FIG.  328.  —  Vector  diagram 

of  E.M.F. 


It   is  best  for  one   starting  the   subject   of   Alternating 
currents,  to   do  the   examples  by  all  three   methods,  using 

1  Unless  otherwise  stated,  the  wave  form  of  an  alternator  is  assumed 
to  be  that  of  a  sine  curve. 


440 


ELEMENTS  OF  ELECTRICITY 


(1)  to  get  a  general  idea  of  the  different  phase  relations, 
etc.,  (2)  and  (3)  to  obtain  correct  mathematical  results. 

Example.  What  is  the  instantaneous  value  of  an  alternating 
E.M.F.  when  it  has  reached  the  45°  phase  of  its  cycle?  The 
maximum  value  is  600  volts. 


360° 


FIG.  329.  FIG.  330. 

Method  (1)  Method  (2) 

Method  (3) 

e  =  E  sin  <j>, 
e= 600  sin  45° 
=  600  X. 707  =  424  volts. 

Solve  as  per  Example  by  Three  Methods 

Problem  1-15.  The  maximum  value  of  an  alternating  E.M.F. 
is  1200  volts.  What  is  the  instantaneous  value  when  65°  of 
the  cycle  have  been  completed? 

Problem  2-15.  What  is  the  instantaneous  value  of  E.M.F. 
of  Problem  1-15  when  it  has  reached  the  200°  phase?  (Note  by 
method  (1)  and  (2),  e  is  seen  to  have  the  same  numerical  value 
as  at  the  200°— 180°,  or  20°  phase.  The  sign,  however,  is  seen 
to  be  negative.) 

Problem  3-15.  The  instantaneous  value  of  an  alternating  E.M.F. 
is  400  volts  at  75°.  What  is  the  maximum  value? 

Problem  4-15.  The  value  of  an  alternating  E.M.F.  is  250  volts 
at  35°.  What  is  it  at  135°? 

Problem  5-15.  Plot  the  sine  curve,  to  some  convenient  scale 
for  one  complete  cycle  for  Problem  4-15. 

Problem  6-15.  The  maximum  value  of  an  alternating  E.M.F. 
is  600  volts.  What  are  the  instantaneous  values  at  the  following 
phases:  20°,  80°,  130°,  210°,  300°,  340°? 


ALTERNATING  CURRENTS  441 

255.  Average  Value  of  an  Alternating  E.M.F.  Since 
half  of  the  instantaneous  values  of  an  alternating  E.M.F. 
are  negative  and  half  are  positive,  and  since  the  negative 
values  are  exactly  equal  to  the  positive  values,  the  average 
of  a  complete  cycle  of  values  must  be  zero.  But  the  actual 
average  value  is  not  zero.  It  would  be  as  reasonable  to 
say  that  the  actual  average  value  of  the  pressure  exerted 
by  the  piston  of  the  water  pump  of  Fig.  309  is  zero,  just 
because  the  pressure  alternates  equally  in  each  direction. 
The  actual  average  value  in  both  cases  is  the  average  of  all 
instantaneous  values  regardless  of  signs,  which  only  indicate 
direction. 

The  average  value  of  an  alternating  E.M.F.  can  be  found 
very  easily  by  plotting  a  number  of  the  instantaneous 
values  throughout  the  cycle  and  finding  their  average. 
Or  the  average  may  be  computed  from  the  equation  by 
calculus.  The  results  of  both  of  these  methods  show  that 
the  average  value  of  an  alternating  E.M.F.  is  .636  times 
the  maximum  value.  In  the  form  of  an  equation,  this 
may  be  written, 

Average  e=.636#. 

\ 

The  average  of  an  alternating  E.M.F.  is  little  used  exeept 
to  compute  the  maximum  value. 

Example.  What  is  the  maximum  voltage  generated  in  a  drum 
armature  consisting  of  300  series  conductors  in  each  path  which 
has  a  speed  of  1200  R.P.M.  Each  conductor  cuts  twice  through 
a  field  of  1,500,000  lines  of  force  during  each  revolution. 

lines  cut  per  sec. 
Average  e  =  —     -^— 

'' _  1,500,000X2X1200X300 

108X60 
=  180  volts. 
But  Average  e  =  .636  E. 


442  ELEMENTS  OF  ELECTRICITY 

Then  E 

.636 

=-i|=283  volts, 
.boo 

Therefore  E  =  283  volts. 

Problem  7-15.  A  2-pole  generator  with  a  drum  armature 
has  a  speed  of  2400  R.P.M.  The  field  has  20,000,000  lines  of 
force.  Number  of  series  conductors  in  each  path  on  armature 
is  500.  What  is  the  maximum  value  of  the  E.M.F.  generated? 

Problem  8-15.  What  is  the  instantaneous  value  of  the  E.M.F. 
at  the  60°  phase  in  Problem  7-15? 

Problem  9-15.  A  4-pole  generator  has  a  drum  wound  arma- 
ture with  2000  series  conductors  in  each  path.  Speed  is  1200 
R.P.M.  Flux  in  each  pole  is  24,000,000  lines.  What  is  average 
value  of  E.M.F.? 

Problem  10-15.  What  is  maximum  value  of  E.M.F.  of  Prob- 
lem 9-15? 

Problem  11-15.  What  is  the  instantaneous  value  of  the  E.M.F. 
of  Problem  9,  at  45°  phase?  Would  the  45°  phase  of  the  E.M.F. 
be  reached  at  the  instant  in  which  the  armature  had  turned  45° 
from  its  zero  or  neutral  position? 

256.  Computation     of    Current    in    A.C.    System.     The 

diagrams,  Figs.  316-330,  which  have  been  drawn  to  show  the 
relation  between  instantaneous  values  of  alternating  E.M.F. 
and  the  maximum  value,  serve  just  as  well  to  show  the 
relation  between  instantaneous  values  of  alternating  current 
and  the  maximum  value.  We  have  merely  to  put  i'i,  i2, 
/,  is,  i*,  —i\i  etc.,  in  the  place  of  e\,  e2,  etc.,  and  all  diagrams 
show  current  instead  of  voltage  values. 

The  same  three  methods  can  be  used  to  solve  problems  in 
current  values  that  we  used  for  E.M.F.  values. 

(1)  Thus  in  Fig.  331,  which  is  identical  with  Fig.  316 
except  for  the  lettering,  i\  is  the  instantaneous  current  at 
the  phase  30°  from  the  zero,  i2  the  value  of  the  current  at 
60°,  and  7  at  90°,  etc.  The  current  thus  follows  a  curve  of 
the  same  shape  as  that  of  the  E.M.F. 


ALTERNATING  CURRENTS 


443 


(2)  The  instantaneous  values  can  also  be  represented  by 
a  vector  or  clock  diagram  as  in  Fig.  332,  which  is  identical 
with  Fig.  317  except  for  the  lettering.  /  represents  the 
maximum  value  of  the  current,  and  i\  represents  the  instan- 
taneous value  of  the  current  when  it  is  at  the  30°  phase; 
ii  is  equal  to  /  sin  30°,  and  can  be  found  from  the  general 
equation : 

i—l  sin(f>; 

i  =  instantaneous  value  of  an  alternating  current; 

/=  maximum  value; 

<f>=  phase  angle  in  degrees. 


210°  240°  270°  300°  390°  SCO" 


A 


0        aOJ    00°     90"   120° 


M80 


\ 


I- 


FIG.  331.— Sine  curve  of  current.     See  Fig.  316.         Fi«.  332.— Vector  diagram 

of  current.     See  Fig.  318. 

As  in  the  case  of  the  alternating  E.M.F.  it  is  best  to  use 
all  three  methods  for  solving  each  problem,  until  the  ideas 
of  each  are  thoroughly  mastered. 

Problem  12-15.  What  is  the  instantaneous  value  of  an  alter- 
nating current  at  20°?  The  maximum  value  is  45  amperes. 

Problem  13-15.  What  is  the  maximum  value  of  an  alternating 
current  when  the  value  at  65°  is  14  amperes? 

Problem  14-15.  What  are  the  instantaneous  values  of  current 
in  Problem  13-15  at  the  180°  phase;  200°;  300°? 

257.  Average  Value  and  Effective  Value  of  Alternating 
Currents.  As  the  average  value  of  an  alternating  E.M.F. 
equals  .636  of  the  maximum  value,  so  likewise,  we  may 


444  ELEMENTS  OF  ELECTRICITY 

write  the  equation  for  the  average  value  of  an  alternating 
current : 

Average  i  =  .636  7. 

Thus  if  the  maximum  value  of  an  alternating  current 
is  50  amperes,  the  average  value  is  50  X. 636  =31. 8 
amperes. 

The  average  value  is  of  little  use,  however,  as  we  rate 
an  alternating  current  on  its  EFFECTIVE  VALUE. 

An  alternating  current  has  no  unit  of  its  own,  but  is 
measured  in  terms  of  the  unit  of  direct  current,  the  AMPERE. 
Now  an  ampere  is  defined  as  that  steady  rate  of  flow  which 
will  deposit  a  standard  amount  of  silver  from  a  standard 
solution  in  one  hour.  But  an  alternating  current  is  not 
a  steady  current,  and  neither  will  it  deposit  any  silver  from 
a  solution;  since  whatever  it  deposits  during  one-half 
a  cycle  it  takes  off  the  next  half,  when  it  is  flowing  in  the 
reverse  direction. 

Accordingly,  in  order  to  compare  the  alternating  with 
the  direct  current,  we  must  use  some  other  property  which 
both  kinds  of  current  possess.  The  most  natural  is  the 
heating  effect  of  each. 

Therefore  an  alternating  current  is  said  to  be  equivalent 
to  a  direct  current  when  it  produces  the  same  average 
heating  effect,  under  exactly  similar  conditions.  This 
value  is  called  the  EFFECTIVE  VALUE  of  an  alternating 
current,  and  is  measured  in  amperes.  It  is  somewhat 
greater  than  the  average  value,  being  equal  to  .707  of  the 
maximum  value.  This  is  because  the  heating  effect  of 
a  current  depends  upon  the  average  of  the  squares  of  each 
instantaneous  value  of  the  current.  The  effective  value 
of  an  alternating  current  is  thus  often  called  the  "  Square 
root  of  the  mean  squares/'  because  it  can  be  found  by 
squaring  a  number  of  the  instantaneous  values  in  a  cycle, 
finding  the  average  of  these  squares,  and  then  extracting 
the  square  root. 


ALTERNATING  CURRENTS  445 

For  all  practical  purposes  the  effective  value  of  an  alter- 
nating current  may  be  found  from  the  following  equation: 


or 


where     lf~  effective  value  of  an  alternating  current; 
/=  maximum  value. 

258.  Effective    Value    of    Alternating    E.M.F.     In    the 

same  way,  the  value  of  alternating  E.M.F.,  by  which  the 
voltage  is  rated,  is  the  EFFECTIVE  value,  and  is  found  by 
the  equation: 

E 


where      Ef=  effective  value  of  alternating  E.M.F.; 
E=  maximum  value. 

When  no  special  value  is  designated,  the  EFFECTIVE  value 
is  always  understood,  for  both  current  and  voltage. 

Proof.  That  the  effective  value  of  an  alternating  current  equals 
the  "  square  root  of  the  average  squares."  v> 

Suppose  two  calorimeters  are  so  arranged  that  one  measures 
the  heat  generated  per  hour  by  an  alternating  current  flowing 
through  a  resistance  R.  The  other  measures  the  heat  generated 
per  hour  by  a  direct  current  flowing  through  an  equal  resistance  R. 

Let  the  two  currents  be  so  regulated  that  the  amount  of  heat 
generated  per  hour  by  each  current  be  equal. 

Then  by  definition: 


where  7j  =  Direct  current  in  amperes; 

//=  effective  value  of  alternating  current. 

The  heat  generated  by  the  direct  current  equals, 

#  =  .24/rf2/ft; 
The  heat  generated  by  an  alternating  current  equals, 

ff  =  .24(av.£)/tt. 
where  i  =  instantaneous  value  of  the  current. 


446  ELEMENTS  OF  ELECTRICITY 

But  the  heat  generated  by  the  direct  current  equals  the  heat 
generated  by  the  alternating  current. 
Therefore, 

.24  (av.  i2}Rt; 


Id  =  \   av.  i2. 

But  h-  If, 

Thus, 


7/=\  av.  i2. 


Effective  value  =  \/Average  of  squares  of  instantaneous  values. 

Since  the  effective  value  of  voltage  and  current  is  always  a 
definite  fractional  part  of  the  maximum  values,  the  EFFECTIVE 
value  may  be  used  in  VECTOR  DIAGRAMS.  To  use  the  effective 
value  this  way  is  merely  equivalent  to  reducing  the  scale  of  the 
diagram. 

Example.  What  is  the  effective  value  of  an  alternating  cur- 
rent whose  maximum  value  is  48  amperes? 

//=. 707X48  =  33.9  amperes. 

Problem  16-16.  The  effective  value  of  an  alternating  current 
is  25  amperes.  What  is  the  greatest  instantaneous  value  of  this 
current? 

Problem  16-16.  What  is  the  average  value  of  the  current  in 
Problem  15-15? 

Problem  17-16.  The  effective  value  of  an  alternating  current 
is  200  amperes.  What  is  the  instantaneous  value  of  30°  phase? 

Problem  18-15.  What  is  the  effective  value  of  an  alternating 
E.M.F.  when  the  instantaneous  value  at  150°  is  500  volts? 

Problem  19-16.  What  is  the  effective  E.M.F.  in  Problem 
7-15? 

Problem  20-16.  What  is  the  effective  voltage  in  Problem 
9-15? 

259.  Phase  Relations  of  Current  and  Voltage.  When- 
ever an  alternating  E.M.F.  is  allowed  to  send  a  current 
through  a  circuit,  of  course  this  current  is  an  alternating 
current.  The  curve  and  equation  representing  the  current 


ALTERNATING  CURRENTS 


447 


will  have  the  same  form  as  the  curve  and  equation  for  the 
E.M.F.,  as  explained  above.  The  vector  diagrams  will 
be  similar  also.  Accordingly,  the  curves  for  the  alter- 
nating voltage  and  current  in  a  circuit  can  both  be  drawn 
on  the  same  pair  of  axes,  as  in  Fig.  333.  Their  vector 
diagrams  also  can  both  be  drawn  on  one  pair  of  axes  as  in 
Fig.  334. 

There  are  three  relations  possible  between  the  current 
and  E.M.F.: 

(1)  Current  and  voltage  curves  may  be  "  in  phase." 
Figs.  333  and  334. 


E.M.F. 


•  FIG.  333. — Sine  curve  of  E.M.F.  and  current. 

(2)  Current    curve  may   "  lag "    behind  voltage    curve. 
Figs.  337  and  338. 

(3)  Current    curve   may    "  lead "  voltage    curve.      Figs. 
340  and  341. 

260.  (I)  E.M.F.  and  Current  in  Phase.  The  current 
and  voltage  may  be  "  in  phase  "  as  in  Figs.  333  and  334. 
That  is,  they  both  are  zero  at  the  same  instant,  both  pass 
through  their  maximum  values  at  the  same  instant,  and, 
in  fact,  are  in  the  same  phase  throughout  their  entire 
cycles.  This  is  the  case  when  the  inductance  and  capacity 
in  the  circuit  are  balanced,  or  when  the  circuit  contains 
resistance  only. 

Fig.  333  represents  the  relations  of  the  current  and  voltage 
when  they  are  "  in  phase."  The  heavy  line  represents 


448 


ELEMENTS  OF  ELECTRICITY 


Fro. 


334.  -Vector   dia- 

wrent  and  volt- 


the  current.  Note  that  both  the  curves  are  at  zero  at  the 
same  instant,  and  that  they  reach  their  maximum  values 
E  and  7  at  the  same  instant  of  phase.  The  value  e\,  at 
the  40°  phase,  corresponds  to  the  value 
i\  at  the  40°  phase  and  is  the  same 
fractional  part  of  E  that  i\  is  of  7. 

Fig.  334  also  shows  the  conditions 
or  relations  of  E.M.F.  to  current  when 
they  are  "  in  phase."  The  lines  7 
and  E  coincide  in  direction  at  all 

same   angle    <£,   from   the    zero  axis. 

The  instantaneous  value  e  is  the  same 
fractional  part  of  E  that  the  instantaneous  value  i  is  of  7. 
This  fraction  is  represented  by  the  expression  "sin  <p  ". 
That  is, 

i  =7  sin  <j> 

and 

e=E  sin  <j>. 

261.  Ohm's  Law  in  A.C.  Circuits.  There  is  always 
likely  to  be  some  confusion  about  the  use  of  Ohm's  law, 
when  the  student  reaches  alternating  current  work.  This 
confusion  disappears  when  one  understands  that  Ohm's 
law  deals  with  VOLTAGE,  CURRENT  AND  RESISTANCE  only, 
and  always  holds  true  as  far  as  the  relative  values  of  these 
three  alone  are  concerned.  The  law  says  nothing  about 
inductance  or  capacity  and  is  not  concerned  with  the  effect 
of  these.  Accordingly  the  student  may  be  sure  that  as 
far  as  the  RESISTANCE  alone  is  concerned,  the  voltage 
required  to  force  a  certain  current  through  a  certain  resist- 
ance is  always  the  product  of  the  current  times  the  resistance, 
whether  the  voltage  is  direct  or  alternating. 

If  some  other  factor  like  a  counter  E.M.F.  due  to  induct- 
ance, etc.,  is  in  the  circuit,  allowance  must  be  made  for  it 
according  to  some  other  law,  not  Ohm's  law.  Therefore 


ALTERNATING  CURRENTS  449 

do  not  have  any  doubts  about  the  validity  of   Ohm's  law. 
It  always  applies  in  so  far  as  RESISTANCE  affects  the  results. 

Example.  A  circuit  containing  resistance  only  has  an  alter- 
nating E.M.F.  impressed  upon  it,  of  which  the  effective  value  is 
110  volts.  If  the  resistance  is  a  lamp  of  55  ohms,  what  is  the 
current  through  the  lamp,  when  the  voltage  is  at  the  30°  phase? 

Solution: 

Ef=.7Q7  E, 

#  =  ^—  =  155  volts. 

Since  there  is  resistance  only  in  the  circuit,  the  current  will  be 
in  phase  with  the  voltage  and  will  be  at  a  maximum  value  when 
the  voltage  is  at  a  maximum. 

Thus  by  Ohm's  law, 

'i 

=-—  =  2.82  amperes. 
oo 

The    instantaneous  value    at    the    30°  phase,  drawn  to    scale, 
as  in  Fig.  335,  =1.4  amperes. 
Or  by  the  equation: 

i  —  l  sin  0 

=  2.82  sin  30 

=  2.82  X.  500  =  1.41  amperes. 

Problem  21-15.  The  average  voltage  in  an  alternating  current 
circuit  containing  20  ohms  resistance  (no  capacity  and  no  induct- 
ance) is  125  volts.  What  is  the  average  current? 

Problem  22-15.  What  is  the  effective  current  in  Problem 
21-15? 

Problem  23-15.  When  the  instantaneous  voltage  in  Problem 
21  is  120  volts,  what  is  the  current  value? 

Problem  24-15.  At  what  phase  would  the  voltage  of  Prob- 
lem 21  be  120  volts? 

Problem  25-15.  Solve  by  vector  diagram.  The  maximum 
voltage  is  200  volts,  resistance  =  2  ohms.  What  is  current,  when 
the  voltage  is  at  the  200°  phase? 


450 


ELEMENTS  OF    ELECTRICITY 


Problem  26-15.     Coil  has  200  turns,   Fig.  336;  resistance,  32 
ohms;  R.P.M.  =  2400;  <£  =  1,000,000. 
(a)  What  is  average  voltage? 
(6)         "       effective  voltage? 
(c)         "       maximum  voltage? 

Problem  27-15.  (a)  How  many  degrees  from  zero  position 
will  coil  in  Problem  26  be,  when  instantaneous  current  equals 
6  amperes?  (b)  What  will  voltage  be  at  this  point? 


FIG.  33o. — Vector  diagram. 


FIG.  336. — Single  coil  generator. 


262.  (II.)  Lagging  Current.  The  current  may  "lag" 
behind  the  voltage  as  in  Figs.  337  and  338.  Note  in  Fig.  337 
that  the  current  curve  is  50°  behind  the  voltage  curve ;  that 


Fro.  337. — Sine  curve  of  current  lagging  50°  behind  the  E.M.F. 

is,  the  voltage  curve  has  a  value  e\  at  50°  while  the  current 
has  a  value  of  zero  at  50°.  Also,  the  voltage  curve  reaches 
its  maximum  at  90°  point  while  the  current  curve  reaches 
it  at  the  140°  point,  50°  later.  So  when  the  voltage  curve 


ALTERNATING  CURRENTS 


451 


has  become  zero  again  at  180°,  the  current  curve  has  but 
just  passed  through  the  maximum  value  and  still  has  a 
value  of  i2. 

Fig.  338  shows  how  this  same  phase  difference  is  repre- 
sented by  a  vector  diagram. 

Since  the  current  reaches  its  maximum  value  50°  later 
than  the  voltage,  the  vector  /  lags  50°  behind  the  vector 
E.  Thus  when  the  vector  E  has  reached  65°,  the  vector  / 
has  reached  only  15°.  The  instantaneous  values  at  this 
instant  are  e  and  i. 


and 


i  -I  sin  15°  or  /  sin  (60°-  50°) 


FIG.  338. — Vector  diagram  of  current 
lagging  50°  behind  the  E.M.F. 


Fl«.  339. — Vector  diagram  for  deter- 
mining the  instantaneous  value  of  the 
current  when  the  E.M.F.  is  at  the  85° 
phase. 


The  general  equations  for  a  "  lagging  "  current  are 


e=E  sin  <£, 
i=I  sin   <>  — 


where 


^=  phase  of  the  voltage  in  degrees; 
6=  difference  in  phase  between  E  and  7. 


Example.  In  an  inductive  A.C.  circuit  the  current  lags  20° 
behind  the  voltage.  The  maximum  value  of  the  current  is  45 
amperes.  What  is  the  instantaneous  value  of  the  current,  when 
the  voltage  is  at  the  85°  phase? 

Solution.  Draw  to  scale  a  vector  diagram  as  in  Fig.  339.  The 
value  of  i  when  voltage  is  at  85°  equals  41  amperes. 


452 


ELEMENTS  OF  ELECTRICITY 


Or  by  the  equation : 


i=I  sin  (0-0); 
=45  sin  (85° -20°) 
=  45  sin  65° 
=40.8  amperes. 

Problem  28-15.  What  instantaneous  value  will  the  current 
and  voltage  have  in  above  example  when  the  voltage  is  at  the 
110°  phase? 

Problem  29-15.  The  phase  difference  in  an  inductive  A.C. 
circuit  is  50°.  The  current  has  an  instantaneous  value  of  25 
amperes  when  the  voltage  is  at  its  maximum,  (a)  What  is  the 
maximum  value  of  the  current?  (6)  What  is  the  effective  value 
of  the  current? 

Problem  30-15.  (a)  What  is  the  maximum  value  of  the  voltage 
(Problem  29-15),  if  the  instantaneous  value  is  500  volts,  when  the 
current  is  at  its  maximum?  (6)  What  is  the  average  value  of 
the  voltage? 

Problem  31-15.  The  maximum  value  of  an  alternating  voltage 
is  1500  volts,  the  maximum  value  of  the  current  is  80  amperes. 
If  the  instantaneous  value  of  the  current  is  25  amperes,  when  the 
instantaneous  value  of  the  voltage  is  800  volts,  what  is  'the  "  phase 
difference  "  between  current  and  voltage? 


E.M.F. 


Current 


360'      30* 


may 


FIG.  340. — Sine  curve  of  current  leading  E.M.F.  by  30°. 

263.  (III.)  Leading  Current.     The    current    curve 
"  lead  "  the  voltage  curve  as  in  Figs.  340  and  341. 

Notice  that  the  current,  Fig.  340,  has  reached  a  value 
ii,  at  30°  while  the  voltage  is  still  zero.  The  current  thus 
"  leads "  the  voltage  by  a  "  phase  difference "  of  30°. 


ALTERNA  TING  C URRENTS 


453 


Also,  while  the  current  reaches  its  maximum  at  90°  the 
voltage  does  not  reach  its  maximum  until  120°;  again  a 
difference  of  30°.  And  so  on  throughout  all  phases  of  the 
cycle. 

In  the  vector  diagram  Fig.  341,  the  current  vector  7  is  30° 
ahead  of  the  voltage  vector  E,  so  that  when  vector  /  has 
moved  55°  from  the  X  axis,  vector  E  has  gone  but  25°. 
The  equations  for  the  instantaneous  values  at  this  instant 
would  be: 

e=#sin25°; 

i=I  sin  55°  or  7  sin  (25° +30°). 
The  general  equations  for  a  "  leading  "  current  are: 

e=E  sin  $, 

i=I  sin  (0  +  0), 
where  0=  phase  difference  between  E  and  7. 


FIG.  341. — Vector  diagram  of 
current  leading  E.M.F.  by 
30°. 


FIG.  342. — Vector  diagram  for  determining 
value  of  current  when  E.M.F.  is  zero.  Cur- 
rent leading  by  40°. 


Example.  In  a  certain  A.C.  circuit  the  current  leads  the  voltage 
by  40°.  The  effective  current  is  100  amperes.  What  is  the  instan- 
taneous current  when  the  voltage  is  0? 

Solution.     Find  maximum  value  of  current  from  equation: 
7/=.7077; 

' 


—  - 


=  141  amperes. 


Draw    the  vector    diagram  to    scale  as  in  Fig.  342.     i  meas- 
ures 91  amperes, 


454  ELEMENTS   OF  ELECTRICITY 

Or  by  the  equation: 


i=141  sin  (0°  +  40°); 
=  90.7  amperes. 

Problem  32-15.  What  will  be  the  instantaneous  value  of  the 
current  in  above  example  when  the  voltage  is  at  a  maximum? 

Problem  33-15.  The  angle  of  lead  of  current  is  80°.  The 
maximum  current  is  56  amperes.  What  .is  current  value  when 
voltage  is  at  135°? 

Problem  34-15.  What  is  value  of  voltage  in  Problem  33, 
when  current  is  maximum,  if  average  voltage  is  250  volts? 

\v  •  264.  Cause  of  a  Lagging  Current.  Inductive  Reactance. 
We  have  seen  in  Chapter  X  that,  if  an  electric  circuit  con- 
tains inductance,  an  E.M.F.  is  induced  in  it  whenever  there 
is  a  change  in  the  current  flowing  in  the  circuit.  This 
induced  E.M.F.  always  opposes  the  change  in  the  current. 

Now  in  an  A.C.  circuit  we  have  a  current  which  is  always 
changing,  so,  if  there  is  any  inductance  in  the  circuit  we 
must  always  have  an  E.M.F.  induced  opposing  this  change. 
When  the  current  is  rising  in  value,  the  induced  E.M.F. 
opposes  this  rise  and  the  value  of  the  current  flowing  is  less 
than  the  "  impressed  voltage  divided  by  the  resistance/' 
just  as  the  current  in  a  motor  which  is  running  is  less  than 
the  impressed  voltage  divided  by  the  resistance  of  the  motor, 
on  account  of  the  opposing  E.M.F.  generated  in  the  rotating 
armature.  Accordingly,  the  rise  of  current  in  the  A.C. 
line  takes  place  later  than  the  rise  in  the  impressed  voltage. 
This  causes  the  current  curve  to  lag  behind  the  voltage 
curve  on  the  rising  side  of  the  curve. 

But  when  the  current  is  decreasing,  the  induced  E.M.F. 
opposes  the  decrease  and  tends  to  keep  the  value  of  the 
current  greater  than  the  impressed  E.M.F.  alone  would. 
Thus  the  current  falls  off  later  than  the  impressed  voltage. 
This  causes  the  current  curve  to  lag  behind  the  voltago 
curve  on  the  descending  side.  Thus  the  current  curve 
"  lags  "  behind  the  voltage  curve  throughout  the  entire 


A LTERNA TING  CURRKXTX 


455 


cycle.  If  a  circuit  contains  nothing  but  the  inductance 
(that  is,  if  the  resistance  and  capacity  are  negligible)  the 
current  lags  90°  behind  the  voltage. 

This  is  evident,  if  'we  consider  that  the  only  voltage 
required  would  be  that  necessary  to  overcome  the  induct- 
ance. This  inductance  would  be  greatest,  not  when  the 
current  was  greatest,  but  when  the  current  was  changing 
at  the  fastest  rate.  By  inspecting  Fig.  343,  it  is  seen  that 
the  current  is  changing  in  value  most  rapidly  as  it  passes 
through  the  zero  value.  Thus  the  voltage  is  greatest  when 
the  current  is  zero.  But  the  voltage  is  always  greatest 


E.M.F. 


Current 


FIG.  343. — Sine  curves  showing  E.M.F.  leading  current  by  90°.     Circuit  contains 
inductance  only. 

when  it  is  at  its  own  90°  phase.  The  current  then  must 
be  zero  when  the  voltage  reaches  its  90°  phase.  The 
current  therefore  lags  90°  behind  the  voltage. 

Again,  the  voltage  must  be  least  when  the  current  is 
changing  at  its  slowest  rate-.  The  current  changes  at  the 
slowest  rate  when  it  has  a  maximum  value.  The  voltage, 
then,  is  zero  when  the  current  is  maximum. 

Thus  the  current  and  voltage  always  differ  in  phase  by 
90°,  which  is  the  difference  in  phase  between  all  maximum 
and  minimum  values.  Whenever  the  voltage  is  zero,  the 
current  is  maximum,  and  whenever  the  voltage  is  maximum 
the  current  is  zero.  Fig.  344  shows  the  vector  diagram 
for  voltage  and  current  in  a  circuit  containing  inductance 
onlv. 


456 


ELEMENTS,  OF  ELECTRICITY 


Since  simple  resistance  in  a  circuit  does  not  cause  a 
current  to  "  lag  "  or  "  lead/'  we  can  say  that  inductance 
has  an  effect  which  is  90°  or  at  right  angles  to  the  effect 
of  resistance.  The  result  of  this  will  be  seen  later  when  the 
combined  effect  of  inductance  and  resistance  is  taken  up. 
Of  course,  if  there  is  also  -resistance  or  capacity  in  the  same 
series  circuit,  the  angle  of  lag  will  not  be  as  great  as  90°. 

Fig.  345  is  the  conventional  way  of  representing  a  circuit 
containing  inductance  only. 


FIG.  344. — Vector  diagram   when   our-          FIG.   345.  —  Conventional   diagram   of 
rent  lags  90°.  circuit  containing  A.O.  generator,  G, 

and  inductance,  L. 


IS 


The  current  equation  for  a  circuit  having  inductance  only 
i=7sin  (0-90°). 


This  opposition  of  self  inductance  to  the  flow  of  the 
current  is  called  INDUCTIVE  REACTANCE,  and  is  measured 
in  OHMS  just  as  RESISTANCE  is.  Also,  just  as  the  current 
flowing  through  a  resistance  is  found  by  the  equation 


Current 


Voltage 
Resistance' 


so  the  current  flowing  through  an  INDUCTIVE  REACTANCE 
is  found  by  the  equation 


Current 


Voltage 
Reactance ' 


ALTERNATING  CURRENTS  457 

or 

7=1,   /    5; 

XL  XL 

where  /  =  current  in  amperes; 

XL=  inductive  reactance  in  ohms; 
E=  voltage  necessary  to  force  current  through  induc- 
tive reactance. 

It  must  be  remembered  that  if  we  use  a  maximum  value 
of  voltage  we  get  a  maximum  value  of  current;  if  we  use 
the  effective  value  of  voltage,  we  get  the  effective  value 
of  the  current;  if  we  use  the  average  value  of  the  voltage, 
we  get  the  average  value  of  the  current;  if  we  use  any 
instantaneous  value  of  the  voltage,  we  get  an  instantaneous 
value  of  the  current  90°  behind  in  phase. 

The  current  value  found  in  this  way  is  always  90°  behind 
the  voltage  producing  it. 

Example.  In  an  A.C.  circuit  containing  40  ohms  inductive 
reactance  (no  resistance  or  capacity),  the  effective  voltage  is 
220  volts,  (a)  What  is  the  effective  current?  (6)  What  is  the 
maximum  current?  (c)  What  is  the  current  when  the  voltage 
is  at  140°  phase? 

Effective  voltage 

Solution.    Effective  current  =  -  =-—  —  , 

Reactance 

or 


220 
(a)  7/=-—  -  =  5.5  amperes. 

and 

7/=.7077; 


707 

(&)  =-^  =  7.8  amperes. 

/  Maximum  current  =  7.8  amperes. 


458 


ELEMENTS  OF   ELECTRICITY 


Draw  vector  diagram  to  scale  as  in  Fig.  346,  when  E.M.F.  is  at 
140°  phase,  then  by  measurement 

7  =  7.8  amperes; 
{=6  amperes; 


or  by  the  equation 


=  I  sin  (0-90) 
=  7.8  sin  (140° -90°) 
=  7.8  sin  50° 
=5.97  amperes. 


E=  220  Volts 


e=? 


I  =  7.8  Amps. 


FIG.  346. — Vector  diagram  of  E  at  140°  and  /  lagging  90°. 

Problem  35-15.  The  average  voltage  in  an  A.C.  circuit  con- 
taining 20  ohms  inductive  reactance  is  500  volts.  Find:  (a) 
Effective  current,  (b)  Instantaneous  value  of  voltage  when 
current  is  15  amperes. 

Problem  36-15.  The  maximum  value  of  the  current  in  an  A.C. 
circuit  containing  inductance  only  is  28  amperes.  The  average 
value  of  the  voltage  is  220  volts.  What  is  the  inductive  react- 
ance? 

Problem  37-15.  At  what  phase  will  the  voltage  be,  in  Problem 
36,  when  the  instantaneous  value  of  the  current  is  10  amperes? 

Problem  38-15.  What  effective  voltage  is  required  to  force 
a  maximum  current  of  20  amperes  through  8  ohms  of  inductive 
reactance? 

Problem  39-15.  What  instantaneous  value  will  voltage  of 
Problem  38  have  when  the  current  is  at  its  maximum  value  of 
20  amperes? 

Problem  40-15.  In  an  A.C.  circuit  containing  inductive  react- 
ance only,  the  voltage  is  1200  and  current  45  amperes.  What 


ALTERNA  TING  C URRENTS 


459 


is  the  instantaneous  value  of  the  current  when  the   voltage  is 
300  volts? 

265.  Cause  of  a  Leading  Current.  Capacity  Reactance. 
In  Chapter  XI,  it  was  stated  that  capacity  in  a  circuit  acts 
like  an  air  chamber  in  a  pump  circuit;  it  tends  to  oppose 
any  change  in  pressure,  and  keep  the  pressure  constant. 
Thus  condensers  in  a  circuit  might  be  thought  of  as  reser- 
voirs, in  which  electricity  is  being  stored  at  the  instants 
when  the  pressure  is  rising.  These  reservoirs  or  condensers 
then  use  the  pressure  of  the  stored  charge,  to  maintain 
the  current  at  the  instants  when  the  pressure  is  dying  out. 
A  current  will  thus  be  flowing  into  a  condenser,  until  it 


E.M.F. 


Current 


FIG.  347. — Sine  curves  showing  Current!  leading  voltage  by  90°.     Circuit  contains 

capacity  only. 

is  fully  charged  and  then  cease.  As  long  as  the  voltage 
across  the  condenser  is  rising,  current  will  be  flowing  into 
the  condenser,  and  the  most'  current  will  be  flowing,  when 
the  voltage  is  rising  most  rapidly.  The  voltage  is  rising 
most  rapidly  when  it  is  at  zero  value  as  shown  in  Fig. 
347.  Thus  the  current  must  be  greatest,  when  the  voltage 
is  zero,  that  is,  the  current's  phase  must  be  90°  when  the 
voltage  phase  is  zero.  The  current  thus  "  leads "  the 
voltage  by  90°.  In  the  same  way  the  current  flows  out 
from  the  condensers  at  the  greatest  rate  when  the  voltage 
is  falling  at  the  fastest  rate.  The  voltage  falls  fastest  at 
270°,  where  it  is  zero.  Thus  current  becomes  a  maximum 
at  this  point,  and  is  still  90°  ahead  of  the  voltage. 


460 


ELEMENTS  OF  ELECTRICITY 


Thus  if  an  A.C.  circuit  contains  capacity  only,  the  current 
"  leads"  the  voltage  90°.  Fig.  348  shows  the  vector 
diagram  of  a  current  leading  the  voltage  by  90°. 

Fig.  349  is  the  conventional  way  of  representing  an  A.C. 
circuit  containing  capacity  only,  the  current  equation  for 
which  is 

i=I  sin  (0+90°). 
Of   course,   if  there   is  resistance   or  self  inductance  in 


series  with  the  capacity,    the 
be  90°. 


angle    of   lead  "    will   not 


FIG   348. — Vector  diagram  when  cur- 
rent leads  voltage  by  90°. 


FIG.  349. — Conventional  diagram  of  A.C. 
circuit  containing  generator,  G,  and 
capacity,  C. 


The  opposition  which  capacity  offers  to  the  flow  of  the 
current  is  called  CAPACITY  REACTANCE,  and  is  measured 
in  Ohms  just  as  resistance  is.  The  symbol  for  capacity 
reactance  is  xc,  and  the  equation  for  voltage  and  current 
relations  in  a  circuit  containing  a  capacity  reactance  of 
xc,  is  similar  to  Ohm's  law  for  the  current-voltage  relations 
in  a  circuit  containing  a  resistance  R.  Thus  we  have  the 
equation : 

Voltage 


Current  = 


or 


Capacity  Reactance7 


,-*,     /,  =  *',  etc. 


ALTERNATING  CURRENTS  461 

where  7=  current  through  capacity  reactance; 
xc=  capacity  reactance  in   ohms; 
E=  voltage  necessary  to  force  current  through  capac- 
ity reactance. 

In  using  the  equation,  the  maximum  value  of  the  current 
must  be  used  with  the  maximum  value  of  the  voltage;  the 
effective  current  with  the  effective  voltage,  etc. 

It  must  be  remembered  that  the  current  value  in  this 
equation  is  always  90°  ahead  of  the  voltage  value. 

Example.  The  voltage  across  a  capacity  reactance  of  4  ohms 
is  120  volts.  What  is  the  current  through  reactance? 


120 

=  —  -—  =  30  amperes. 
4 

Example.     What  is  the  maximum  value  of  the  current  in  above 
example? 

7/=.707/; 


.707 

30 
=  ——  =  42.4  amperes. 

A  vector  diagram  for  this  example  is  shown  in  Fig.  350.     Note 
that  the  maximum  value  of  the 
current  (/)  is  90°  ahead    of    the 
maximum  value  of  the  voltage  E. 

Problem  41-15.  Draw  vector 
diagram  for  above  example  with 
voltage  at  15°  phase  and  find 
instantaneous  values  of  current 
and  voltage. 

Problem  42-16.     A   110  volt,        FlQ.  350._Veotordiagram;  current 
A.C.  circuit  contains  a  condenser  leading  by  90°'. 

only.     If  the  current  is  2  amperes, 
what  is  the  capacity  reactance  of  the  condenser? 


462  ELEMENTS  OF  ELECTRICITY 

Problem  43-15.  When  the  instantaneous  voltage  in  Problem 
42  is  120  volts,  what  instantaneous  value  will  the  current  have? 

Problem  44-15.  What  voltage  is  necessary  to  force  a  maximum 
current  of  20  amperes  through  a  circuit  containing  50  ohms  of 
capacity  reactance? 

Problem  45-15.  (a)  Draw  vector  diagram  and  determine 
instantaneous  value  of  voltage  when  instantaneous  current  in 
Problem  44  is  5  amperes.  (6)  In  what  phase  will  voltage  be  at  that 
instant? 

266.  Computation  of  Inductive  Reactance.  In  taking 
up  the  computation  of  Inductive  Reactance,  that  is,  the 
opposition  offered  to  the  flow  of  an  alternating  current  by 
an  inductance,  it  is  necessary  to  have  well  in  mind  the 
meaning  of  the  term  INDUCTANCE  and  its  measurement. 

Inductance  has  been  denned  in  Chapter  X  as  the  inertia 
reaction  to  any  change  in  the  current  flowing  in  an  electric 
circuit. 

There  is  one  unit  of  INDUCTANCE,  that  is,  one  HENRY, 
when  a  change  of  one  AMPERE  per  sec.  induces  an  E.M.F. 
of  one  VOLT. 

This  is  expressed  by  the  equation  : 


where  average  E=  average  voltage  induced  in  circuit; 
L=inductance  of  circuit  in  henrys; 
7=  change  in  current  in  amperes; 
t  =time  of  change  in  seconds. 

This  equation  may  be  stated  in  words  as  follows  : 

The  voltage  induced  in  any  circuit  by  change  in  the  current 

is  equal   to  the  product  of  the  inductance  in  henrys  times  the 

rate  of  change  in  amperes  per  sec. 

Now   in   an   alternating   current,   during   one   cycle,   the 

current  makes  the  following  four  changes,  as  per  Fig.  351. 


ALTERNATING  CURRENTS 


463 


(1)  The  current  rises  from  zero  to  maximum. 

(2)  falls  from  maximum  to  zero. 

(3)  "  rises  from  zero  to  maximum  in  opposite 

direction. 

(4)  "  falls  from  this  maximum  to  zero. 

In  each  of  the  four  changes,  the  current  makes  a  complete 
change  from  zero  to  the  maximum  value  7,  or  vice  versa, 
in  one  direction  or  the  other.  The  time  of  each  complete 
change  can  be  found  as  follows : 

The  number  of  cycles  per  sec.  =/. 

The  time  for  one  cycle  =.—  sec. 

each  change  =J  time  for  each  cycle. 


270°  360° 


90°          180' 


FIG.  351.  —  Sine  curve  of  current  during  one  complete  cycle. 

The  rate  of  change,  then,  of  an  alternating  current,  must 


be  /  amperes  in—  -  sec.,  or  (4/7)  amperes  per  sec. 

4/ 
Thus  the  expression, 

—  (rate  of  change  of  current)  must  equal  4/7. 
The  expression 


becomes 


ave.  E=L  —  \ 


ave.  E=4fLL 


464  ELEMENTS  OF  ELECTRICITY 

That  is,  the  average  voltage  induced  in  an  inductive  circuit 
equals  four  times  the  product  of  the  frequency  by  the  induct- 
ance by  the  maximum  current. 

If  a  circuit  contains  nothing  but  inductance,  the  impressed 
voltage  merely  has  to  overcome  the  induced  voltage,  and 
is  therefore  equal  to  it;  just  as  in  starting  an  unloaded 
machine  which  has  no  friction,  an  impressed  force  has  to 
overcome  merely  the  inertia  reaction  and  must  be  equal 
to  it. 

Thus  the  average  impressed  voltage  must  equal  the  average 
induced  voltage.  The  maximum  voltage  E  is  generally 
used  in  the  equation,  and  this  is  found  as  follows: 

average  #=4/L7, 

but  average  E  =  .636#, 

then  .636  #  =4  /L7, 

or  E  =6.28  f  LI. 

Since  6.28  =2?r 

the  equation  is  generally  written 

E=2xfLL 

This  is  the  general  equation  for  the  voltage  necessary  to 
send  an  alternating  current  through  an  inductance,  that 
is,  through  an  inductive  reactance. 

But  we  have  seen  that  the  voltage  necessary  to  send  an 
alternating  current  through  an  inductive  reactance  was 
found  by  the  following  equation: 


XL 

or  E=IXL. 

But  E=2nfLl, 

then  IXL=2nfLI- 

and  XL=2xfL. 


ALTERNATING  CURRENTS  466 

Therefore  the  value  of   an  Inductive  reactance  in  Ohms 
is  equal  to  2?r  times  the  Frequency  times  the  Inductance. 

Example.     What  is  the  inductive  reactance  of  an  A.C.  circuit 
of  .08  henry  inductance,  if  the  frequency  is  60  cycles  1 


=  2X3.14X60X.08 
=  30.2  ohms. 

Problem  46-15.  What  current  will  flow  in  circuit  of  above 
example  if  the  voltage  is  110  volts? 

Problem  47-15.  Draw  vector  diagram  for  Problem  46,  when 
instantaneous  value  of  the  current  is  25  amperes. 

Problem  48-15.  The  inductance  of  an  A.C.  circuit  is  .2  henry, 
the  voltage  is  110  volts.  At  what  frequency  will  the  current  be 
3  amperes? 

Problem  49-15.  What  will  be  the  reactance  of  circuit  in  Prob- 
lem 48-15? 

Problem  50-15.  A  550  volt  A.C.  circuit  has  a  frequency  of 
133  cycles.  What  inductance  must  be  placed  in  the  circuit  to 
limit  the  current  to  40  amperes? 

Problem  51-15.  What  instantaneous  value  will  current  in 
Problem  50  have  when  the  voltage  is  at  the  40°  phase? 

Probelm  52-15.  A  coil  40  cms.  long,  containing  85  turns 
of  wire,  is  wound  on  a  wrought  -iron  core  15  cms.  in  diameter 
(j«  =  1000).  The  wire  is  of  such  size  that  the  resistance  is  prac- 
tically zero.  If  this  coil  were  put  directly  across  a  110  volt  A.C. 
circuit  of  a  frequency  of  60  cycles,  what  current  would  flow 
through  coil?  See  equation  in  Chapter  X  for  computing  induct- 
ance of  a  coil. 

267.  Computation  of  Capacity  Reactance.  In  a  circuit 
containing  capacity,  such  as  a  condenser,  the  current  charges 
and  discharges  the  condenser  four  times  each  cycle,  as  seen 
from  Fig.  351,  which  represents  the  current  curve  in  such 
a  line: 

(1)  Charging  condenser  (positively). 

(2)  Discharging  condenser. 

(3)  Charging  condenser   (negatively). 

(4)  Discharging  condenser. 


466  ELEMENTS  OF  ELECTRICITY 


The  time  consumed  for  each  complete  charge  or  dis- 
charge, then,  must  be  —  sec.  The  quantity  of  charge  in  a 
condenser  can  be  found  by  the  equation  (Chapter  XI) 


where  q=  charge  in  coulombs; 

7?  =  voltage  across  terminals  of  condenser; 
C=  capacity  or  (number  of  coulombs  per  volt). 

But  the  quantity  of  water  or  electricity  or  anything  that 
flows,  must  equal  the  average  rate  of  flow  times  the  time 
of  flow;  thus 

q=Av.  I  XL 
But 

K  ^  .....  s  <  in 

Therefore 

!       Av-  7 


Since  also 

then 

Fr_Av.J. 

*/ 
or 

Av.  7 
h  ~  4/C  J 
but 

Av.  7  =.6367. 


Therefore 


.6367  _      7      _/._/!    V 

-*~*  ^    /» y^»  ^»     r-k  r-»   /» X^i  <r-*         /» y^|  I     Q         /» X~)     I  "*   ' 


4/C      6.28/C 
The  usual  form  of  the  equation  is 


ALTERNATING  CURRENTS  467 

The  law  may  be  stated  in  words  as  follows:  The  voltage 
necessary  to  cause  an  alternating  current  to  flow  through 
a  line  containing  capacity  only,  equals  the  product  of  the 
current  times  the  reciprocal  of  2x  times  the  frequency  by  the 
capacity. 

But  we  have  seen  that  the  voltage  to  force  a  current  to 
flow  in  a  circuit  containing  capacity  only,  also  equals: 


where  Xc=the  capacity  reactance. 

Therefore 


and 


-- 

c       27T/C* 

Thus  the  value  of  Capacity  Reactance  in  Ohms  equals  the 
Reciprocal  of  2n  times  the  Frequency  times  the  Capacity. 

Example.     What  is  the  capacity  reactance  of  an  A.C.  circuit 
containing  35  microfarads  capacity  if  the  frequency  is  GO  cycles? 


1 


6.28X60X35X10-6 
1 


.0132 
=  75. 8  ohms. 

Problem  63-15.  What  current  will  550  volts  force  through 
circuit  of  above  example? 

Problem  54-15.  What  instantaneous  value  will  voltage  in 
Problem  53-15  have  when  the  instantaneous  current  is  8  amperes? 

Problem  55-15.  Find  the  average  value  of  the  voltage  across 
a  condenser  of  20  rnfs.  capacity,  if  a  current  of  3  amperes  flows, 
when  the  frequency  is  133  cycles. 


468  ELEMENTS  OF  ELECTRICITY 

Problem  56-15.  What  will  the  instantaneous  current  value 
be  in  Problem  55,  when  the  voltage  is  at  80°  phase? 

Problem  57-15.  The  voltage  in  a  25~A.C.  circuit  is  550  volts. 
If  a  condenser  of  the  following  dimensions  be  placed  directly 
across  the  line,  what  current  will  flow  in  circuit?  2000  plates 
of  lead  foil  80  cms.  by  60  cms.  Dielectric  of  waxed  paper  .008 
cm.  thick.  See  Chapter  XI  for  equation  for  capacity  of  con- 
denser. 

Problem  58-16.  Three  condensers  of  20,  80,  and  50  mfs.  capac- 
.ity  respectively,  are  placed  in  parallel  across  a  60  cycle  A.C. 
circuit  of  220  volts.  What  is  maximum  current  in  circuit? 

268.  Reactance,  a  Combination  of  Inductive  and  Capac- 
ity Reactance.  Inductive  reactance  and  capacity  reactance 
have  exactly  opposite  effects  upon  the  phase  relation  of 
the  current  curve  to  the  voltage  curve. 

Inductive  reactance  causes  the  current  to  "  lag." 
Capacity  reactance  causes  the  current  to  "  lead." 
Thus  when  the  inductive  and  capacity  reactance  are  com- 
bined in  series,  they  tend  to  neutralize  each  other.     The 
combined  effect  then  is  the  difference  between  them,  and  is 
called  the  REACTANCE.     The  REACTANCE  of  such  a  series 
circuit  is  found  by  subtracting  the  capacity  reactance  from 
the  inductive  reactance.    This  is  expressed  by  the  equation : 


X=XT-X. 


c> 


where      X = reactance  in  ohms; 

XL  —  inductive  reactance  in  ohms  ; 
Xc=  capacity  reactance  in  ohms. 

When  the  result  is  positive,  the  reactance  is  due  to  a  ±arger 
inductive;  when  negative,  to  a  larger  capacity  reactance. 

Example.     What  is  the  reactance  of  a  circuit  containing   14 
ohms  inductive  reactance  in  series  with  5  ohms  capacity  reactance? 

14-5  =  9  ohms  reactance  (inductive). 


ALTERNATING  CURRENTS  469 

Example.  What  is  the  reactance  of  a  60  cycle  A.C.  circuit 
containing  an  inductance  of  1.5  Henrys  in  series  with  a  capacity 
of  40  microfarads? 


=  6.28X60  XI. 5 
:  565  ohms. 
1 


1 

"6.28X60X40X10-^ 

=  T0151 
=  66.3. 
X  =  XL—XC 
=  565-66 
=  499  ohms  reactance. 

Problem  59-15.  What  is  the  reactance  of  a  133  cycle  A.C. 
circuit,  which  contains  20  mfs.  capacity  in  series  with  4  Henrys 
inductance? 

Problem  60-15.  What  current  flows  in  circuit  in  Problem  59, 
if  the  voltage  is  550  volts? 

Problem  61-15.  What  current  will  flow  in  a  60  cycle,  220  volt 
circuit  containing  2  Henrys  inductance  and  25  mfs.  capacity  in 
series? 

Problem  62-15.     What  current  will  flow  in  circuit  of  Problem 
61,  if    frequency  falls    to  30    cycles,  but 
everything  else  remains  unchanged? 

Problem  63-15.  Draw  vector  diagram 
for  Problem  61,  and  determine  value  of 
current  when  voltage  is  at  30°  phase. 

Problem  64-15.    If  coil  B  of  Fig.  352,     fla  3S2._inductio^ 
containing  2  henrys  inductance,  and  con-  with  iron  core, 

denser  of   Problem  57-15  were  placed  in 

series  in  a  500  volt  60  cycle  circuit,  what  would  be  the  maximum 
current  flowing  in  circuit? 

269.  Impedance,  a  Combination  of  Resistance  and 
Reactance.  In  a  circuit  containing  RESISTANCE  only, 


470 


ELEMENTS  OF  ELECTRICITY 


the  current  is  in  phase  with  the  voltage.  In  a  circuit  con- 
taining REACTANCE  only,  the  current  either  "  leads  "  or 
"  lags  "  90°.  The  effect  of  reactance  on  the  phase  of  the 
current  is  thus  at  an  angle  of  90°  to  the  effect  of  resistance. 
The  result,  therefore,  of  a  series  combination  of  the  two 
cannot  be  found  by  simple  addition  or  subtraction.  It  can 
easily  be  found,  by  the  graphical  method  as  follows: 

Suppose  it  were  required  to  find  the  result  of  a  series 
combination  of  4  ohms  RESISTANCE  and  3  ohms  inductive 

REACTANCE.  Draw  horizontal 
line  R  to  scale  to  represent  4 
ohms  resistance  (Fig.  353) .  Then, 
since  the  effect  of  reactance  is 
at  right  angles,  90°,  to  resistance, 
draw  the  line  XL  to  scale  at 
right  angles  from  end  of  R,  to 
represent  3  ohms  reactance.  The 

FIG.   353. — Graphical   method    of      ..  __  . 

finding  impedance  from  resist-     Ime   Z   then  represents  the  IM- 

ance  and  inductive  reactance. 

PEDANCE,  or  resultant   of  R  and 

XL.  In  this  case,  when  scaled  off,  Z-=5  ohms,  which  is 
neither  the  sum  of  4  and  3  nor  the  difference. 

Since  the  lines  R,  XL  and  Z  always  form  a  right  triangle, 
of  which  Z  is  the  hypothenuse,  an  equation  can  be  formed, 
from  which  the  value  of  Z  can  be  found  directly,  without 
drawing  the  lines  to  scale.  In  every  right  triangle  "  the 
square  of  the  hypothenuse  equals  the  sum  of  the  squares 
of  the  other  two  sides." 
Thus 


and 


In  above  problem, 


=  \/16+9 

-V25 

=  5  ohms. 


ALTERNATING  CURRENTS 


471 


If  the  reactance  is  a  capacity  reactance,  it  is  customary 
to  draw  the  diagram  as  in  Fig.  354,  with  the  line  representing 
the  capacity  reactance,  Xc,  drawn  downward  to  show  that 
it  acts  in  the  opposite  direction  to  inductive  reactance, 
which  is  drawn  upward.  The  mathematical  result  and 
the  equation  are  the  same  as  in  the  case  of  inductive  react- 
ance. 


If  the  circuit  contains  both  inductance  and  capacity  in 
series   with  the    resistance,   then   the    impedance    can    be 


XL=9 


X=3 


FIG.  354. — Graphical  method  of  finding  imped- 
ance from  resistance  and  capacity  reactance. 


FIG.  355. — Graphical  determi- 
nation of  impedance. 


found  by  first  subtracting  the  capacity  reactance  Xc  from 
the  inductive  reactance  XL,  and  combining  the  result,  X, 
with  the  resistance,  R  as  per  Figs.  353  and  354.  The  equa- 
tion would  then  become : 


Fig.  355  shows  the  graphical  method  of  obtaining  the 
impedance  in  a  series  circuit  of  resistance,  inductance  and 
capacity. 

Suppose  9  ohms  inductive  reactance  and  6  ohms  capacity 
reactance,  are  in  series  with  4  ohms  resistance. 

The  resistance  R  is  represented  by  the  horizontal  line 
R.  The  inductive  reactance  XL  by  the  line  XL  drawn  up 


472 


ELEMENTS   OF  ELECTRICITY 


from  end  of  R,  at  right  angles  to  R.  The  capacity  reactance 
is  represented  by  the  line  Xc  drawn  down  from  end  of  R, 
at  right  angles  to  R.  The  resultant  of  XL  and  Xc  is  found 
by  subtracting  Xc  from  XL  and  is  equal  to  X.  That  is: 

X  =  X  —X 

=  9-6 
=3  ohms. 


FIG.  355a.  —  Moore  light  installed  in  silk  mill  for  matching  colors. 

The  line  Z  is  then  drawn  as  the  resultant  of  R  and  X. 
Z  by  scale  =5  ohms. 
By  the  equation 


=  V42  +  (9-6)2 


\/25 


When  inductance  and  resistance  exist  in  the  same  piece 
of   machinery   as  in   an   induction  coil,  the  same    current 


ALTERNATING  CURRENTS  473 

must  be  forced  through  both  reactance  and  resistance.  The 
'resistance  and  reactance  are  therefore  considered  to  be 
in  series. 

The  same  applies  to  the  capacity  and  resistance  of  a 
condenser,  etc. 

270.  General  Law  for  A.C.  Circuits.  The  impedance 
of  a  circuit  represents  the  total  opposition  to  flow  of  the 
current.  Thus  the  relation  between  voltage  and  current 
can  be  stated  simply  as  follows  : 

The  CURRENT  flowing  in  an  alternating  current  circuit 
equals  the  quotient  of.  the  VOLTAGE  divided  by  the 
IMPEDANCE.  This  is  sometimes  called  Ohm's  law  of  the 
alternating  current.  It  is  represented  by  the  equation 


Z' 

where   7  =  maximum,  average  or  effective  current  in  amperes  ; 
E=  respectively    maximum,     average,    or     effective 

voltage  in  volts. 
Z  =  impedance  in  ohms. 

Example.  What  current  flows  in  an  A.C.  circuit  of  110  volts 
and  4  ohms  impedance? 

<4 

=  —  —  =  22.5  amperes. 
4 

Example.  A  v  series  circuit  contains  an  induction  coil  of  10 
ohms  resistance,  6  ohms  reactance,  and  a  condenser  of  25  ohms 
capacity  reactance,  negligible  resistance  .  If  voltage  is  220  volts, 
what  current  flows? 

The  reactance  of  the  induct  ion  coil  must  be  inductive,  thus: 


=  6-25 

=  —  19  ohms  reactance. 


474  ELEMENTS  OF  ELECTRICITY 


=  V  102  +  (-19)2 

=  \/100  +  361 

=  V46T 

=21.5  ohms  impedance. 

=t 

220 


21.5 
=  10.24  amperes. 

Problem  65-15.  What  is  the  impedance  in  a  circuit  containing 
25  ohms  inductive  reactance,  in  series  with  70  ohms  capacity 
reactance  and  40  ohms  resistance? 

Problem  66-15.  What  current  will  flow  in  circuit  of  Problem 
65  if  the  voltage  is  550  volts? 

Problem  67-15.  A  60  cycle  110  volt  circuit  has  12  ohms  resist- 
ance in  series  with  .2  henry  inductance  and  24  mfs.  capacity. 
What  is  impedance  of  the  circuit? 

Problem  68-15.     What  is  the  maximum  current  in  Problem  67? 

Problem  69-15.  How  much  would  impedance  of  circuit  be, 
if  the  frequency  changed  to  120  ~? 

Problem  70-15.  What  would  be  the  current  in  circuit  of 
Problem  69? 

271.  Voltage  and  Current  Relations  in  Series  and  Paral- 
lel  Circuits.     The  relations  existing  between  voltage  and 
current  in  series   and  parallel    DIRECT  CURRENT    circuits 
have  been  explained  in  Chapter  III. 

The  same  method  of  computing  these  relations  will 
apply  to  alternating  current  circuits.  Care  must  be  taken, 
however,  to  take  into  account  the  effect  of  any  difference 
in  phase  between  the  voltage  and  current. 

272.  Series    Circuits.     Voltage    and    Current    Relations. 
The  current  flowing   in  a   series   circuit   is   of    course   the 
same  throughout,  whether  it  be  D.C  or  A.C. 


ALTERNATING  CURRENTS  475 

The  VOLTAGE  across  a  series  combination  equals  the 
-VECTOR  SUM  of  the  voltages  across  the  separate  parts. 

This  is  exactly  analogous  to  the  rule  for  direct  currents; 
the  sum  of  the  voltages  in  direct  current  being  the  algebraic 
sum,  since  the  voltage  and  current  are  always  in  phase. 

The  meaning  and  application  of  the  expression  VECTOR 
SUM  will  be  made  clear  by  a  study  of  the  following  illustra- 
tions. 

Assume  a  series  circuit  (Fig.  356)  to  be  made  up  of  a 
reactance  x,  and  a  resistance  R. 
Let      E  =IZ  =  volt  age  across  the  combination. 
EI  =IX  =  voltage  across  the  reactance. 
E2=  IR=  voltage  across  the  resistance. 


FIG.  356. — Series  circuit  containing  resistance  and  capacity  reactance. 

Electromotive  force  E  does  not  equal  the  algebraic  sum  of 
EI  and  E2,  because  of  the  effect  of  the  reactance  on  the 
phase  relations  between  current  and  voltage. 

Yet  the  electromotive  force  E  equals  the  vector  sum  of 
the  two  electromotive  forces  E2  and  EI,  in  exactly  the  same 
manner  that  a  single  mechanical  force  equals  the  vector  sum 
of  its  two  components,  which  are  at  an  angle  to  each  other. 
Thus  the  force  F  (Fig.  357),  is  equal  to  the  VECTOR  SUM 
of  its  two  components  FI  and  F2.  If,  therefore,  we  repre- 
sent the  two  electromotive  forces  E1  and  E2  by  lines 
drawn  at  the  proper  phase  angles  to  each  other,  the  electro- 
motive force  E  will  be  represented  by  their  vector  sum. 

In  Fig.  358  OA  is  drawn  to  scale  equal  to  IR=E2.  OB 
is  drawn  at  right  angles  to  OA  equal  to  IX  =  E\.  The 
resultant  or  vector  sum  of  OA  and  OB  is  OC,  equal  to  IZ  =E. 

OB  is  drawn  at  an  angle  of  90°  to  OA ,  because  the  effect 
of  reactance  is  always  at  90°  angle 'to  the  effect  of  resistance. 


476 


ELEMENTS  OF  ELECTRICITY 


In  Fig.  354  the  line  Xc  is  drawn  at  right  angles  to  the  line 
R  for  the  same  reason. 

By  scaling  off  the  line  OC,  or  E,  we  can  find  the  value 
of  the  voltage  across  the  series  combination  X  and  R. 

Or  the  value  of  E  may  be  computed  from  the  equation 
for  the  right  triangle 

E2=E12+E22  (since  line  0£=line  AC). 


Furthermore,   the   angle    0   is   equal   to   angle   of   phase 
difference    between    voltage    and    current    in    the    circuit. 


FIG.  357.  —  Graphical  solution  for 
resultant  of  two  forces  at  right 
angles  to  each  other. 


FIG.  358. — Graphical  solution  for  voltage 
across  series  combination  of  resistance 
and  reactance. 


This  will  be  apparent  on  consideration  of  the  following 
points : 

Resistance  alone  does  not  throw  the  current  out  of  phase 
with  the  voltage.  Therefore  the  current  must  have  the 
same  phase  as  the  component  IR  or  E2,  which  is  merely 
that  component  of  the  voltage  across  the  resistance  R. 

Since  the  direction  of  E  represents  the  direction  of  the 
voltage  across  the  combination,  and  the  direction  of  E2 
the  direction  of  the  current,  the  angle  between  E  and '  E2 
must  equal  the  angle  of  phase  difference  between  voltage 
across  the  combination  and  the  current  through  the  com- 
bination. 


ALTERNATING  CURRENTS 


477 


If  X  is  inductive  reactance,  the  angle  6  represents  an 
angle  of  "lag,"  or  the  number  of  degrees  the  current  is 
behind  the  voltage. 

If  X  is  capacity  reactance,  the  angle  6  represents  an 
angle  of  "  lead,"  or  the  number  of  degrees  the  current  is 
ahead  of  the  voltage. 

This  angle  may  either  be  measured  or  be  ascertained  by 
means  of  the  following  equation: 

AC 


l  =tan  <9; 

#2 

El=IX  and  E2=1R, 
El     IX    X 


but 

Thus 
or  since 

then 


Knowing  the  value  of  X  and  R,  the  tangent  of  the  angle 
6  can  be  computed,  and  the  angle  6  found  by  reference 
to  a  table  of  tangents. 

Example.  In  a  series  A.C.  circuit  consisting  of  8  ohms  induct- 
ive reactance  and  10  ohms  resistance  a  current  of  6  amperes  is 
flowing,  (a)  W7hat  is  the  volt- 
age across  each  part  of  the  cir- 
cuit? (b)  What  is  the  voltage 
across  the  combination?  (c) 
What  is  the  difference  in  phase 
between  the  current  and  voltage 
in  the  combination? 

Solution,  (a)  /X- 6X8  =  48 
volts  across  inductive  reactance 
X.  IR  =  6X10-60  volts  across 
resistance  R.  Draw  IR  to  scafe 
horizontally  (Fig.  359).  Draw 
IX  to  scale  vertically. 

Then  E/  (the  resultant  voltage  of  IR  and  IX)  to  scale  =  7.68 


Fir,.  359. — E  equals  effective  voltage 
across  X  and  R. 


478 


ELEMENTS  OF  ELECTRICITY 


volts.     Angle  6,  the  angle  of  difference  in  phase  between  voltage 
and  current  in  circuit,  measures  39°.     By  computation 


=V64  +  100 
=Vl64. 
=  12.8  ohms. 
Ef=IfZ 
=6X12.8 
=  76.8  volts. 

tan  0=^p-8; 

.8  =  tan  38°  40'; 
0  =  38°  40'. 

The  current  lags  38°  40'  behind  the  voltage,  since  X  is 
INDUCTIVE  reactance. 

The  vector  diagram  for  current  and  voltage  would  there- 
fore be  like  that  of  Fig.  360. 

The  diagram  of  Fig.  361  shows  the  phase  relation  of 
current  and  voltage  by  means  of  sine  curves  for  current  and 
voltage. 

When  voltage  has  reached  the  instantaneous  value  e\, 

at  the  39°  phase,  the  current 
<76.s  has  a  value  of  zero.  The  cur- 
rent reaches  its  maximum  value 
7,  39°  later  than  the  voltage 
reaches  its  maximum  value  E. 
The  current  curve,  then,  is  every- 
where approximately  39°  behind 
the  voltage  curve. 

Note    the    fact,    startling:    to 

FIG.  360. — yector  diagram  of  cur- 
rent  lagging    38°  40'  behind   the     students    wllO      have    Worked     On 
voltage. 

D.C.  circuits  only,  that  the  alge- 
braic sum  of  the  voltages  across  the  separate  parts  (48  + 
60  =  108)  is  greater  than  the  actual  voltage  across  the 
combination  (76.8  volts). 


=6.0 


ALTERNATING  CURRENTS 


479 


THE  COMBINED  RESISTANCE  of  a  number  of  separate  resist- 
ances in  series  equals  the  algebraic  sum  of  the  separate 
resistances. 

THE  COMBINED  REACTANCE  of  a  number  of  reactances  in 
series  equals  the  algebraic  sum  of  the  separate  reactances. 
(Inductive  reactances  being  positive  and  capacity  react- 
ances negative.) 

THE  IMPEDANCE,  or  combined  effect  of  series  resistance 
and  reactance,  has  been  dealt  with  earlier  in  the  chapter. 


oltage 


yoc 


180° 


3603/ 


Fi«.  361. — Sine  curves  of  current  lagging  39°  behind  the  voltage. 

THE  COMBINED  IMPEDANCE  of  a  number  of  impedances 
in  series  equals  the  VECTOR  SUM  of  the  separate  impedances. 

Problem  71-15.  Find  the  current  flowing  in  an  A.C.  series 
circuit  containing  16  ohms  resistance  and  20  ohms  capacity  react- 
ance. Voltage  is  220  volts. 

Problem  72-15.  Find  the  voltage  across  each  part  of  circuit  in 
Problem  71. 

Problem  73-15.  Find  the  phase  relation  between  the  current 
and  voltage  in  Problem  71. 

Problem  74-15.  A  550  volt  A.C.  circuit  has  the  following 
pieces  in  series  across  the  line:  (1)  A  resistance  of  14  ohms:  (2) 
a  capacity  reactance  of  20  ohms :  (3)  12  ohms  inductive  reactance. 
(4)  10  ohms  resistance.  (5)  4  ohms  capacity  reactance.  Find: 
(a)  What  current  flows  through  combination,  (b)  Voltage  across 
each  part,  (c)  Phase  relation  of  current  and  voltage. 

Problem  75-15.  The  voltage  across  R,  Fig.  362,  is  20  volts 
A.C.  (a)  Find  voltage  (Ef)  across  each  part  and  across  the 


480 


ELEMENTS   OF   ELECTRICITY 


combination.      (b)  Find    phase  difference   between    current   and 
voltage. 

Problem  76-15.     A  coil  containing  4  ohms  resistance  and  .5 
henry  inductance  is  placed  in  series  with  a  15  mf.  condenser  of 


8  Ohms 


5  Ohms 

^AAAr 


2  Ohms 


FIG.  362. — Series    combination   of   resistance,    inductive   reactance,  and    capacity 

reactance. 

2  ohms  resistance,  across  a  60  cycle  A.C.  circuit  of  550  volts. 
What  current  flows  in  circuit? 

Problem  77-15.     WTiat  is  phase  relation  between  current  and 
voltage  in  Problem  76-15? 

273.  Parallel  Circuits.     Voltage  and   Current  Relations. 

The  VOLTAGE  across  a  parallel  combination  is  the  same  as 

the  voltage  across  each  branch,  as 

in  a  B.C.  circuit.  xL=50hms 

The  CURRENT  through  a  parallel 
combination  is  the  vector  sum  or 
resultant  of  the  currents  through 
each  branch. 


Example.     In  the  parallel  combina- 
tion, Fig.  363,  it  is  desired  to  find  the 
current  flowing  through  each   branch,     Fl?ion3^7^S±i,  i 
and  the  total  current  through  the  com- 
bination. 

120 


reactance  and   capacity  re- 
actance. 


/c  =  — -  =  15  amperes  (current  through  Xc), 


R), 


current  through  combination. 


ALTERNATING  CURRENTS 


481 


Note.  The  sign  (+)  is  to  stand  for  the  VECTOR  SUM  of  the  two 
quantities  between  which  it  is  placed. 

Draw  IR  to  scale,  Fig.  364,  to  represent  the  current  through 
the  resistance  R.  At  right  angles  to  IR  (ahead  of  IR), 
draw  Ic  to  represent  the  current  through  Xc,  leading  the 
current  in  IR  by  90°.  At  right  angles  to  IR  (behind  /#), 
draw  IL  to  represent  the  current  through  7L,  lagging  90° 
behind  current  in  IR.  Construct  the  parallelogram  OACG 
and  draw  the  diagonal  OC.  This  represents  the  resultant 
of  IL  and  IR. 

That  is,  OC=IR@IL. 


FIG.  364.-7Graj>hical  solution  for  the 
current  in  parallel,  combination  of 
Fig.  363. 


Fir,.  365. — Simpler  graphical  solution 
for  current  in  parallel  combination 
of  Fig.  363. 


Construct  a  parallelogram  ODBC,  and  draw  the  diagonal 
OB  or  If.  This  diagonal  If  represents  the  resultant  of  IL 
and  OC.  That  is,  If  =  OC®Ic=IR©IL@Ic. 

Therefore  If  represents  (to  scale)  the  amount  of  the  cur- 
rent in  the  parallel  combination  of  R,  Xc  and  XL.  The 
voltage  must  be  in  phase  with  IR,  the  current  through  R. 
Thus  the  angle  6  between  If  and  IR  is  a  measure  of  the  phase 
difference  between  voltage  and  the  current  of  the  combina- 
tion. In  this  case  0  represents  an  angle  of  lag,  since  the 
inductive  reactance  XL  is  not  entirely  counteracted  by  the 
capacity  reactance  Xc.  By  measurement, 

7y=41  amperes; 
0  =  13°. 


482  ELEMENTS  OF  ELECTRICITY 

Instead  of  Fig.  364,  the  simpler  Fig.  365  may  be  drawn. 
The  line  Ix  represents  the  current  through  the  combined 
reactances,  XL,  and  Xc.  The  resultant  current,  If,  through 
the  whole  combination,  can  then  be  found  by  the  construc- 
tion of  but  one  parallelogram.  Of  course  the  results 
obtained  are  the  same.  The  values  may  be  obtained  by 
computation  as  follows: 


=  \/(24-15)2+402 

=  XX92+402 

=  \/1681 

=41  amps,  (current  through  combination). 


From   tables  of  tangents 

.225=tan  12°  41'. 

274.  Adding   Alternating   Current   Curves.     Currents  in 
the  different  branches  of  a  parallel  circuit  may  be  either 

(a)  in  phase. 
or 

(6)  out  of  phase. 
(a)    In  Phase. 

We  will  consider  first  the  case  where  the  currents  are  in  phase. 
Assume  that  the  currents  7t  and  72  in  the  parallel  circuits  (Fig. 
366),  are  in  phase. 
Then  the  equation 


becomes,  ,_.      T 

1  —  1  i~r  !•>. 

The  curves  for  this  condition  are  represented  by  Fig.  367.  The 
current  flowing  in  the  part  BC  of  the  circuit  is  represented  by  the 
curve  /.  The  current  flowing  in  ADB  is  represented  by  the  curve 


ALTERNATING  CURRENTS 


483 


7t;  the  current  through  A  KB,  by  the  curve  7,.  Note  that  the 
curve  /  is  a  combination  of  curve  7t  and  72.  Both  currents  I \ 
and  7,  do  not  flow  at  once  in  BC,  but  merely  one  current,  their 
resultant  /. 

Any  instantaneous  value  of  /  is  the  sum  of  the  corresponding 
values  of  7:  and  72.     For  instance,  i  =  il  +  it. 
or, 

7  sin  <f>  =  Ii  sin  <£  +  72  sin  <£. 


FIG.  366. — Parallel  circuit. 

(6)     Out  of  Phase. 

We  will  consider  the  following  cases: 

(1)  When  the  current  phases  differ  by  180° 

(2)  "  "  "  90° 

(3)  "  "  "  0°  (general  case), 


Fio.  367. — Current  curve  7  equals  sum  of  sine  curves  /2  and  J\.  h  and  7i  are  in  phase. 

(1)     Currents  180°  Different  in  Phase. 

If  the  current  7t   lags    180°  behind  72    Fig.  368  will  represent 
the  curves. 

The  equation 

7  =  7,07,; 
becomes. 


484 


ELEMENTS   OF  ELECTRICITY 


Any  instantaneous  value,  such  as  i,  equals  i>  —  i,,  but  i,  and  z\ 
are  not  in  the  same  phase.     The  equation  for  instantaneous  value 


is, 
or 


sn  <£  =     sn 


sn 


180°). 


sn      --      sn 


FIG.  368. — Current  curve  7  equals  the  algebraic  sum  of  the  sine  curves  7«  and  7i. 
7j  is  180°  behind  h. 

(2)  Currents  90°  Different  in  Phase.  If  the  current  72 
lags  90°  behind  7lf  Fig.  369  represents  the  curves  of  the  current 
in  the  different  parts  of  the  circuit.  The  equation  for  the  re- 
sultant current  /  in  part  BC  is, 


FIG.  369. — Current  curve  7  equals  the  sum  of  the  sine  curves  and  J\.     Iz  lags  90° 

behind  7i. 

Any  instantaneous  value  can  be  found  by  adding  algebraically 
the  values  of  /i  and  /2  at  the  same  instant. 
Thus  in  Fig.  369 

i  =  ii-i2 

Thus,  if  two  currents  which  differ  in  phase  by  90°  are  added  to- 
gether, the  resulting  current  will  be  the  sum  of  the  instantaneous 


ALTERNATING   CURRENTS 


485 


values  of  the  two  currents,  and  will  not  be  in  phase  with  either 
current.  It  may  be  said  to  differ  in  phase  from  either  current  by 
an  angle  of  6°. 

The  instantaneous  value  can  be  found  from  the  equation: 


=    sn    <- 


sn 


sn    <,- 


FIG   370  — Current  -curve  7  equals  the  sum  of  sine  curves  h  and  7i.     Ii  lags  9° 

behind  7i. 

(3)  Currents  6°  Different  in  Phase.  Let  currents  7t  and  /2 
differ  in  phase  by  0°. 

Then  /,  Fig.  370,  represents  the  sum  of  these  two  currents 
according  to  the  equation, 

/  =  /!©/,. 


FIG.  371. — Vector  diagram  of  currents  in  Fig.  370. 

Since  I2  lags  6°  behind  /i,  the  resultant  of  the  two  will  be  in 
phase  with  neither,  but  may  be  considered  to  lag  0t°  behind  7t. 
The  equation  for  any  instantaneous  value  then  becomes, 

i=/sin  (<£-0i)=/i  sin 


486  ELEMENTS  OF   ELECTRICITY 

Fig.  371  represents  vector  diagram  of  this  case  corresponding 
to  curves  in  Fig.  370. 

Thus  the  current  flowing  in  the  circuit  of  the  above  example 
on  page  480  would  be  represented  by  a  single  curve  7,  Fig.  372, 
made  up  of  curves  7#,  IL  and  7C. 


FIG.  372. — Current  curve  7  equals  sum  of  currents  7C,  /£,  and  7/j».     7^  lags  90° 
behind  IR  and  180°  behind  7C. 

Any  instantaneous  value  on  this  curve  would  be  made  up  of 
corresponding  instantaneous  values  on  the  other  two  curves. 

The  curve  /  also  lags  13°  behind  the  voltage,  which  of  course 
must  be  in  phase  with  curve  IR.  Any  instantaneous  value  may 
be  found  from  the  equation 

i=7  sin  (0-13°)  =7iz  sin  <£+  IL  sin  (0-90°)  +/c  sin  (<£  +  90°). 

275.  THE  IMPEDANCE  of  a  parallel  combination  can 
always  be  found  without  any  special  rule.  Merely  divide 
the  CURRENT  through  the  combination  (found  as  above)  by 
the  VOLTAGE  across  the  combination.  If  the  voltage  is  not 
given,  assume  a  voltage  and  proceed  as  above. 

Example  1.  What  is  the  impedance  of  the  parallel  circuit  of 
above  example,  page  480? 

Er  (across  combination) 

Z  (of  combination)  = ——r- — — - — -, 

//  (through  combination) 

120 

Z  =  — -  =  2.93  ohms, 
41 


ALTERNATING  CURRENTS 


487 


Example  2.  A  circuit  arranged  as  in  Fig.  373  presents  all 
fundamental  combinations,  (a)  Find  the  value  of  the  impedance 
when  the  following  values  are  known.  (6)  What  is  the  difference 
in  phase  between  current  and  voltage? 


XL  =  6  ohms. 
/?  =  4  ohms. 


2  =  8  ohms. 


L2  =  3  ohms. 
R3  =  5  ohms. 


FIG;  373. 


XL2 


To  find  impedance  (ZJ  and  phase  angle  of   branch  composed 
of  XL  and  Rt  in  series,  draw  Fig.  374. 

Zt= 7.2  ohms, 

^  =  56°. 


FIG.  374. — Graphical  solution  for 
impedance  of  series  combina- 
tion XL  and  Ri  in  Fig.  373. 

By  computation: 


FIG.  375. — Graphical  method  of  finding 
impedance  of  series  combination  Rz 
and  Xc;  Fig.  373. 


=\/16+36 

=  \/52~ 

=  7.22  ohms. 


tan      = 


488 


ELEMENTS    OF    ELECTRICITY 


The  impedance  of  branch  composed  of  XL  and  li^  in  series  is 
7.22  ohms. 

The  current  through  this  branch  of  the   circuit  lags  56°  20' 
behind  the   voltage. 

To  find  Impedance  Z2  and  phase  relations  of  branch  composed 
of  #2  and  Xc  draw  Fig.  375. 

By  measurement  : 

Z2  =  13ohms.     02  =  51°. 

By  computation  : 


=\/64  +  100 
=  12.8  ohms. 
tan  02  =  ^0=1.25. 
From  table,  02  =  ol°  20'. 


FIG.  376. — Graphical  method  of  finding  FIG.  377. — Analytical  method  of  find- 
current  through  parallel  combination  ing  current  through  parallel  corn- 
in  Fig.  373.  bination  in  Fig.  373. 

Impedance  of  branch  composed  of  R2  and  Xc  =  12.8  ohms. 
The  current  through  this  branch  leads  the  voltage  by  51°  20'. 
To  find  the  combined  impedance  of  above  branches  in  parallel 
draw  Fig.  376.  By  assuming  a  voltage  of  110  volts  across  AB,  the 
following  current  must  flow: 

Through  Zt, 


T = T22  =  15'3  amperes 


Through  Z  2, 


ALTERNATING  CURRENTS  489 

In  Fig.  376,  E  represents  the  phase  of  the  voltage  E.  Current 
through  Zi  lags  56°  20'  behind  E;  then  the  line,  I\  of  15.3  amperes, 
at  an  angle  of  56°  20'  lag  to  E,  represents  the  current  through  Zr 
Similarly  line  /2  represents  a  current  of  8.1  amperes,  51°  20'  ahead 
of  E  in  phase.  The  line  /,  resultant  of  7j  and  72,  represents  the 
current  through  the  parallel  combination;  and  the  angle  0  repre- 
sents the  difference  in  phase  between  the  current  through  com- 
bination and  voltage  across  combination. 

By  measurement, 

7  =  15  amperes.     0  =  25°. 
By  computation  (referring  to  Fig.  377), 

/  =  /!©/, 

=  V  (/2cos51°  +  /1  cos56)2  +  (/1sin56°-72sin51°):! 


=  V  (5.1  +8.54)2  +  (12.7  -6.3) 
=  V(13.64)2-(6.4)2 


=  V  186+41 

=  V227 

=  15.1  amperes. 


0  =  25°. 

The  current  through  the  parallel  combination  thus  equals  15.1 
amperes  and  lags  25°  behind  the  voltage. 

The  impedance  of  the  parallel  combination  AB  can  now  be 
found  by  the  equation  : 

7-E 

T 

=  -  =  7.3  ohms. 
15.1 

The  impedance  Z  of  the  series  combination  BC  can  be  found  from 
Fig.  378. 
'  By  measurement  : 

Z3  =  5.8  ohms. 


490  ELEMENTS   OF  ELECTRICITY 

By  computation : 


=  V25  +  9 
=  5.84  ohms. 
tan  0  =  £  =  .f>, 


The  impedance  (Z3)  of  the  series  circuit  BC  is  5.84  ohms,  and 
its  effect  is  to  cause  the  current  through  it  to  lag  31°  behind  the 
voltage. 

But  in  series  with  above   impedance    Z3,    is    the    impedance 
Z   of   the    parallel    circuit  AB,  which  causes  the  current  to  lag 


R3=5 


FIG.  378. — Graphical  method  of          FIG.  379. — Graphical  method  of  finding  corn- 
finding     impedance    of    series  bined  impedance  of  Z  and  Zi. 
combination  Ra  and  X^2. 

25°  behind   the   voltage   across   BC.     The   combined   impedance 
of  these  two  (Z  and  Z3)  can  be  found  from  Fig.  379. 
By  measurement : 

Z4  =  13ohms, 

4 

By  computation : 


t  =  V  (Z  sin  25°  +  Z  .  sin  31°)  2  +  (Z  cos  25°  +  Z3  cos  31°)  ; 


=  V172 

=  13.1  ohms. 


ALTERNATING  CURRENTS 


491 


From  table,  04  =  27°40'. 

Thus  the  impedance  of  the  series-parallel  combination  is  13.1 
ohms,  and  the  phase  difference  between  current  and  voltage  is 
27°  40'  lag. 

276.  Summation  of  Voltage,  Current  and  Impedance 
Relations  in  Series  and  in  Parallel  Combinations. 

(1)  SERIES. 

Voltage  across  combination  equals  vector  sum  of  separate 
voltages. 

Current  is  same  throughout  series  combination. 
Impedance  equals  vector  sum  of  separate  impedances. 

(2)  PARALLEL. 

Voltage  is  same  across  each  branch  of  combination. 
Current  equals  vector  sum  of  separate  currents. 
Impedance  equals  (assumed)  voltage  divided  by  the  cur- 
rent which  (assumed)  voltage  forces  through  combination. 


FIG.  380. — Capacities  in  parallel. 

Problem  78-15.     What  current  flows  through  a  circuit  con- 
nected as  in  Fig.  380  under  following  conditions? 

C  has  a  resistance  of  2  ohms. 
C       "    capacity  of  20  mfs. 
Cl     "    resistance  of  1  ohm. 
GI      "    capacity  of  40  mfs. 
£/=110  volts,  25  cycles. 

Problem  79-16.     What  is  difference  in  phase  between  E/  and 
//  of  circuit  in  Problem  78-15? 


492 


ELEMENTS  OF  ELECTRICITY 


Problem  80-15.     What  current  would  flow  in  circuit  of  Prob- 
lem 78-15  if  the  condensers  C  and  C\  had  no  resistance? 
Problem  81-15. 

L  has  4  ohms  resistance  and  .4  henry  inductance,  Fig.  381. 
Ll  has  8  ohms  resistance  and  .8  henry  inductance. 
#  =  110  volts,  25  cycles. 

What  current  will  flow  through  above  circuit? 
What  will  be  the    phase    relations    between  current    //   and 
voltage  #/? 
Problem    82-15.     In  Fig.  382, 

L,=6  henrys; 
L2  =  3  henrys; 
C1  =  40mfs.; 
C,  =  20mfs.; 


FIG.  381.— Inductances  in  parallel. 


R2  =  5  ohms; 
#3  =  3  ohms; 
R+  =  4  ohms. 
E/= 220  volts,  60  cycles. 


Find:  Current  through  AB; 

CD-, 

FG. 
Phase  relation  of  current  through  FG  to  voltage  across  FG. 


Ci 


RI 


Ls          R, 

-^5W>— ^WW-rG 


FIG.  382. 

276.  Power  in  A.C.  Circuits.  The  power  taken  at  any 
instant  by  an  alternating  current  circuit  equals  the  prod- 
uct of  the  voltage  at  that  instant  times  the  current  at 
that  instant.  In  the  form  of  an  equation  it  may  be  written: 

p=ie; 

p  =  instantaneous  power  in  watts; 
e  =  u  voltage  in  volts; 

i=  "  current  in  amperes. 


ALTERNATING  CURRENTS 


493 


The  power  curve  can  then  be  drawn,  each  point  on  which 
shall  be  the  product  of  the  value  of  the  current  times  the 
value  of  the  voltage  at  that  instant. 

The  question  of  power  consumed  in  an  alternating  cur- 
rent circuit  is  divided  into  two  parts: 

(1)  When  the  current  and  voltage  are  IN  PHASE. 

(2)  When  the  current  and  voltage  are  OUT  OF  PHASE. 
(1)   Fig.  383  represents  such  a  curve,  when  the  current 

and  voltage  are  in  phase. 


FIG.  383.- 


-Power  curve  P  equals  product  of  sine  curves  E  and  I.     E  and  T  in 
phase. 


Any  value  as  p,  on  the  power  loop  P,  is  the  product  of 
the  corresponding  value  of  the  voltage  e,  on  the  voltage  loop, 
E,  times  the  current  i,  on  the  current  loop  /. 

Note  that  all  the  power  loops  are  positive  (above  the  axis) , 
even  when  the  current  and  voltage  are  negative.  This 
is  for  the  reason  that  the  signs  of  the  current  and  voltage 
values  change  at  the  same  instant,  and  thus  the  product 
is  always  a  product  of  two  positive  values  or"  of  two  negative 
values;  either  of  which  products  is  always  positive. 

Therefore,  when  the  current  and  voltage  are  in  phase  the 
power  is  all  positive.  That  is,  no  power  is  flowing  backward 
and  being  returned  to  the  generator.  The  effective  power 
in  such  a  circuit,  then,  must  be  the  total  of  the  product 
of  the  effective  voltage  times  the  effective  current.  The 


494  ELEMENTS  OF  ELECTRICITY 

equation  therefore  for  a  non-inductive  circuit,  or  for  a  cir- 
cuit containing  resistance  only,  is 


Another  way  of  considering  this  case  is  to  note  that  there 
being  no  reactance  in  the  circuit  all  the  power  must  be 
expended  in  overcoming  resistance.  The  power  used  to 
overcome  resistance  at  any  instant  must  be  i2R,  or; 


The  average  of  this  power  would  be  Av  i2R  or, 

(1)  P=Avi2R. 

But  by  definition 

VAvi?=If. 
Therefore, 

Avi2=If2; 
and  since 

P=Avi2R; 
then 

(2)  P=lf2R. 

But  by  Ohm's  law,  the  voltage  to  force  a  given  current 
through  a  given  resistance  is 

Ef=IfR. 

Then,  substituting    this  value    for  IfR  in    equations   (2), 
we  have, 

P-W. 

277.  (2)  Power  in  Inductive  Circuits.  Power  Factor. 
Fig.  384  represents  the  power  curve,  when  the  voltage  and 
current  are  6°  OUT  OF  PHASE.  We  have  seen  that  such  a 
case  exists,  when  the  circuit  contains  reactance.  The  same 
equation  for  instantaneous  power  holds  true  —  the  instan- 


ALTERNATING  CURRENTS 


495 


taneous   power  equals   the   product   of    the  current   value 
times  the  voltage  value  at  the  same  instant,  or 

p  =ie. 


FIG.  384. — Power  curve  equals  product  of  current  curve  7  and  voltage  curve  E.     I 

lags  behind  E. 

Any  point  in  the  power  curve  such  as  p  is  the  product  of 
the  instantaneous  values  i  and  e  on  the  voltage  and  cur- 
rent curves. 

But  note  that  the  power  loops  B  and  D  are  negative; 
that  is,  they  are  the  product  of  a  negative  value  of  voltage 
times  a  positive  value  of  current,  or  vice  versa.  The  power 
then  is  not  all  flowing  one  way.  Part  of  it,  represented  by 
the  small  negative  loops  B  and  D,  is  being  returned  to  the 
generator.  The  average  power  used  by  such  a  circuit  is  then 
no  longer  the  entire  product 
of  the  effective  value  of 
voltage  times  the  effective 

value  of  the  current.    If  the     

circuit  contained  resistance 
only,  such  would  be  the 
case,  as  we  have  seen  above. 


IR 


Therefore  that  part  of  the 
power  which  is  used  to  over- 
come resistance  only  in  any 
circuit  must  be  the  power  consumed  in  the  circuit. 


FIG.  385. — Graphical  method  of  resolving 
a  current  If  into  two  components,  Ix  and 
/jg,  at  right  angles  to  each  other. 


496  ELEMENTS  OF  ELECTRICITY 

By  means  of  a  vector  diagram  as  Fig.  385,  we  can  see  what 
fraction  of  the  power  this  must  be  in  each  case. 

Let  lj  represent  the  effective  current  lagging  6°  behind 
the  voltage.  We  may  consider  this  current  If  as  composed 
of  two  components,  IR  and  lx.  The  component  7^  is  in 
phase  with  the  voltage,  and  the  component  Ix  is  at  right 
angles  to  the  voltage. 

Now  since  IR  is  the  only  component  of  current,  to  main- 
tain which  power  is  required,  we  may  say  that  the  average 
power  in  the  circuit  equals  the  effective  voltage  times  the 
component  of  the  effective  current  in  phase  with  voltage. 
The  equation  would  then  be 


but 

IR=If  cos  0; 
therefore 

P=EfIfcos  0. 

This  is  the  general  equation  for  power  consumed  in  any 
electric  circuit. 

The  term  "cos  0"  is  called  the  "  POWER.  FACTOR  "  since 
it  is  the  factor  by  which  the  apparent  power  (amperes 
times  volts)  must  be  multiplied  in  order  to  obtain  the  true 
power. 

When  the  current  is  in  phase  with  the  voltage,  0=0 
and  cos  0  =  1.  The  power  factor  is  then  said  to  be  one',  or 
UNITY. 

The  load  which  is  most  likely  to  produce  a  lagging  cur- 
rent is  that  which  is  composed  of  a  number  of  induction 
motors.  Fig.  385a  represents  a  motor  of  this  type.  It  has 
the  advantage  of  using  no  commutator  or  collecting  rings. 

278.  Wattless  Component  of  Current.  Since  there  is 
no  power  consumed  in  forcing  the  component  of  current 
Ix  (Fig.  385),  through  the  circuit,  this  component  is  some- 
times called  the  WATTLESS  COMPONENT  of  the  current. 


A  L  TERN  A  TING  C  URREN  TS 


497 


The  reason  why  it  requires  no  power  to  force  tne  wattless 
component  through  the  circuit  can  be  understood,  if  we 
consider  that  the  power  taken  by  this  current  is  used 
either  to  charge  a  condenser,  when  of  course  this  charge 
all  flows  back  into  the  generator,  or  to  set  up  a  counter 
electromotive  force  in  a  coil,  which  of  course  helps  send 


FIG.  385a. — Induction  motor. 


Takes  lagging  current  and  reduces   power  factor 
the  lin 


of  the  line. 

the    current   back  toward  the   generator,   as   the    E.M.F. 
reverses. 

279.  Use  of  Ammeter  and  Voltmeter  to  Measure  Power. 
Wattmeter.  Since  A.C.  voltmeters  and  ammeters  indicate 
effective  values,  when  a  voltmeter  and  ammeter  are  used 
to  measure  power,  the  product  of  the  volts  times  the  amperes 
or  volt-amperes  does  not  always  give  the  true  power.  If  the 
current  "  leads "  or  "  lags/'  the  angle  of  lead  (or  lag) 


498 


ELEMENTS  OF  ELECTRICITY 


must  be  known,  so  that  the  power  factor,  (cos  0),  may  be 
used  to  reduce  the  volt-amperes  to  watts. 

A  wattmeter,  however,  is  so  constructed  that  it  indicates 
the  true  power  in  an  A.C  circuit.  Its  readings,  therefore, 
do  not  need  to  be  multiplied  by  the  power  factor  (cos  0) . 

Example  1.  In  a  certain  A.C.  circuit  where  the  current  lags 
30°,  an  ammeter  indicates  20  amperes  and  a  voltmeter  115  volts. 
What  would  a  wattmeter  indicate? 

A  wattmeter  would  read  power  directly,  or 

P  =  20XH5Xcos30° 
=  20X115X.866 
=  2000  watts. 

Example  2.  There  are  5  amperes  flowing  in  an  A.C.  circuit 
under  a  pressure  of  110  volts.  If  the  current  is  in  phase  with  the 
voltage,  what  power  is  consumed  by  the  circuit? 

P  =  //£/  cos  0, 
Here  cos  6  =  1 . 

Therefore,  P  =  //#/. 

Fig.  386  shows  vector  diagram  for  this  case 
=  5X110 
=  550  watts. 


FIG.  386. — Vector  diagram.    Current  and       FIG.  387. — Vector  diagram.     Current 
voltage  in  phase.  lags  25°  behind  voltage. 

If  the    voltage  in   the   above  example   leads   the    current    by 
25°,  what  power  is  being  consumed? 

P  =  //E/cos  6; 
=  5X110Xcos25° 
=  550  X. 905 
=  498  watts. 


ALTERNATING  CURRENTS 


499 


Fig.  387  shows  vector  diagram  for  a  case  where  voltage  leads 
current  by  25°. 

Example  3.  A  110  volt  A.C.  circuit  contains  40  ohms  resistance. 
What  is  the  average  power  in  the  circuit? 

r     Ef 
^; 
—  =  2.75  amperes. 

4\J  •  ^ 

P  =  IfEf 
=  2.75X110 
=  303  watts. 


FIG.  388. — Impedance  Z,  com- 
posed of  resistance  R  and 
reactance  X. 


FIG.  389. — Vector  diagram.  Current  lagging 
90°  behind  voltage.  No  component  in 
phase  with  voltage. 


If  the  circuit  in  above  example  3  contained  10  ohms  resistance 
and  30  ohms  reactance,  what  would  be  the  average  power  in  the 
circuit? 

Referring  to  Fig.  388: 


=A   1000 
=  31.  6  ohms; 


*-» 

_  110 

~3iir 


:  3.48  amperes. 


=  71°  65'; 

=  IfE/cos  0 

=  3.48  XI 10  X.  313 

=  120  watts. 


500  ELEMENTS  OF  ELECTRICITY 

Example  4.  Assume  that  the  40  ohms  in  above  example  3  were 
reactance,  what  power  would  be  consumed  by  the  circuit? 

Reactance  alone  in  a  circuit  causes  the  current  and  voltage 
to  differ  in  phase  by  90°.  There  is  then  no  component  in  phase 
with  the  voltage  as  shown  by  Fig.  389  and  the  power  should  be 
zero.  Fig.  390  indicates  this,  in  that  it  shows  that  the  negative 
power  loops  are  as  large  as  the  positive  power  loops,  when  the 
phase  difference  is  90°.  Thus  all  the  power  is  being  returned  to 
the  generator.  The  equation  shows  that  no  power  is  being  used. 


0  =  90°; 
cos  90°  =  0; 

Therefore,  P  =  I/E/XQ 

=  0. 


FIG.  390. — Power  curve  P  has  as  large  negative  as  positive  loops  when  current  ana 
voltage  differ  90°  in  phase. 

Problem  83-15.  What  power  is  consumed  in  a  circuit  con- 
taining 8  ohms  impedance?  Voltage  =  110  volts.  Power  fac- 
tor =.9. 

Problem  84-15.  What  is  the  power  factor  in  a  circuit  which 
contains  20  ohms  resistance  and  4  ohms  reactance? 

Problem  85-15.  What  power  is  consumed  in  circuit  in  Prob- 
lem 84,  if  the  voltage  is  220  volts? 

Problem  86-15.  How  many  amperes  are  there  in  the  wattless 
current  in  Problem  85-15? 

Problem  87-15.  A  110  volt  60  cycle  circuit  contains  10  ohms 
resistance,  (a)  What  current  flows?  (6)  What  power  is  con- 
sumed? 

Problem  88-15.  How  much  would  current  in  Problem  87  be 
reduced  to,  if  a  coil  of  negligible  resistance  and  .2  henrys  inductance 
were  placed  in  series  with  the  10  ohm  resistance? 


ALTERNATING  CURRENTS  501 

Problem  89-15.  What  power  would  be  consumed  by  circuit 
in  Problem  88? 

Problem  90-15.  A  coil  of  .015  henry  inductance  and  2  ohms 
resistance  is  placed  on  a  25  cycle  110  volt  circuit,  (a)  What  is 
power  factor?  (6)  What  power  is  consumed  by  coil? 

Problem  91-15.  An  ammeter  inserted  in  a  110  volt  A.C. 
circuit  indicates  14  amperes.  Wattmeter  reads  1.4  K.W.  What 
is  power  factor  and  phase  difference  between  current  and  voltage? 

280.  To  Determine  the  Necessary  Brush  Potential  of 
A.C.  Generators.  (1)  When  the  Power  Factor  of  the  Load 
is  Unity.  Assume  required  voltage  V  at  motor  M  =  110 
volts,  current  of  M=40  amperes.  Current  of  M  in  phase 
with  voltage.  Line  resistance  =.2  ohm.  No  reactance  in 
line. 

Solution.  Since  there  is  no  reactance  in  the  circuit,  the  current 
throughout  is  in  phase  with  the  voltage.  The  voltage  across  the 
combination  of  series  resistances  then  equals  the  sum  of  the  sepa- 
rate resistances,  as  in  a  D.C.  circuit. 

Voltage  across  motor  .  =110  volts 

Voltage  lost  in  line  =  .2  X  40      =     8  volts 

Volts  required  at  generator     =118  volts 

Power  delivered  at  generator  to  line  =  118X40  =  4720  K.W. 


Bii 


FIG.  391. — A  D.C.  shunt  generator  feeding  a  bank  of  lamps. 

This  is  exactly  the  same  as  though  a  D.C.  generator,  Fig.  391, 
were  feeding  40  amperes  to  a  bank  of  lamps  at  110  volts  over  a 
line  of  .2  ohm.  The  voltage  of  the  generator  would  have  to  be 
118  volts. 

(2)  When  the  Current  in  Load  Lags.  Assume  same 
power  is  taken  by  motor,  except  that  the  current  through 
motor  lags  so  as  to  produce  a  power  factor  of  .90,  i.e., 
cos  0=.90. 


502  ELEMENTS  OF   ELECTRICITY 

Power  consumed  by  motor  in  (1)  =110X40=4.4  K.W. 
Power  in  motor  in  this  example  must  equal  4.4  K.W.  also. 
But  power  =EI  cos  6.  If  we  assume  that  the  voltage  is 
to  remain  110  volts,  the  current  through  motor  can  be 
found  by  the  above  equation. 

4400  =  110  X/X. 00; 

4440 

7= — —=44.4  amperes. 
«-/y 

That  is,  when  the  current  through  motor  lags,  it  must 
be  larger,  in  order  that  the  power  remain  the  same  as  when 
the  power  factor  was  unity,  voltage  remaining  constant. 

Construct  ve-ctor  diagram  as  per  Fig.  392. 


I  R  =  8.9 
FIG.  392. — Voltage  E  is  vector  sum  of  IR  and  V. 

Draw  line  /  of  indefinite  length  to  represent  current 
phase.  Voltage  to  force  current  through  line  resistance 
of  .2  ohm  equals  8.9  volts  and  is  in  phase  with  current. 
Therefore,  draw  IR  along  /  to  scale  to  equal  8.9. 

P.F.=.90; 
Thus 

cos  0  =  .90; 
cos  24. 5°  =.90. 

Therefore  the  angle  between  voltage  and  current  of  motor 
equals  24.5°. 

Draw  line  V  to  scale  to  equal  110  at  an  angle  of  24.5° 
to  line  I. 

The  brush  potential  E  of  generator,  then,  equals  the  vector 


ALTERNATING  CURRENTS  503 

sum,  or  resultant,  of  IR  and  V.  Construct  parallelogram 
and  scale  off  E. 

By  this  method  E  ==  117  volts,  angle  0\,  between  current 
and  generator  voltage  =23°. 

Power  delivered  by  generator  equals 

P  =  117X44.4  cos  23° 
=  117X44.4X.92 
=4.780  K.W. 

By  computation, 

E=I@RV-, 


sin  24.5)2 


=  V(8.9  +99)2  +  (45.7)2 

=  Vl  1700  +2080 
=  117  volts. 

The  angle  Q\  between  voltage  and  current  of  generator, 
can  be  found  by  the  following  equation. 

V  sin  24.5 


108 
=  .423 
=22.9°; 

Power  =  1  17  X  44.4  X  cos  22.9° 
=  117X44.4X.92 
=4.780  K.W. 


The  power  delivered  by  the  generator,  when  the  current 
is  out  of  phase  with  the  voltage  must  be  more  than  when 
current  is  in  phase,  in  order  to  supply  motor  with  given 
constant  load,  over  line  of  given  resistance, 


504 


ELEMENTS  OF  ELECTRICITY 


(3)  When  the  Load  Consists  of  Parallel  Pieces  with 
Different  Power  Factors.  In  Fig.  393,  M  takes  40  amperes 
at  110  volts.  Power  factor  =85%.  Each  lamp  between 
FB  takes  10  amperes.  Power  factor  of  lamps,  unity. 


.16  Ohm 


.lOhm 


T 


D  46  Ohm  F  .1  Ohm  K 

FIG.  393.— A.C.  generator  feeding  parallel  pieces. 

Find: 

(1)  Current  through  each  section  of  line. 

(2)  Voltage  across  BF. 

(3)  Voltage  across  generator. 

(4)  P.F.  of  generator    load. 

(5)  Efficiency  of  transmission. 

Draw  Fig.  394  similar  to  Fig.  392,  making  IR=8  and  the 
angle  between  V,  voltage  across  motor,  and  /,  current 
through  motor  =32°;  since  cos  32°  =  .85=power  factor. 


FIG.  394. — Graphical  method  of  finding  voltage  across  BF,  Fig.  393. 

The  resultant  of  IR,  8  volts  and  V,  110  volts,  by  measure- 
ment equals  117  volts.  Angle  0\y  between  current  and 
voltage  in  line  BCKF=30°. 

The  current  in  line  A B  and  FD  can  now  be  found  by 
adding  vectorially  currents  of  40  amperes  lagging  30° 
behind  the  voltage  and  of  20  amperes  in  phase  with  volt- 


ALTERNATING  CURRENTS  505 

age,  as  in  Fig.  395.  By  measurement,  current  /  in  AB 
and  DF=5S  amperes.  Angle  between  current  in  line  and 
voltage  of  line  =20°.  Voltage  to  force  this  current  through 
section  of  line  AB  and  ZXF  =  .32X58  =  18.6  volts.  Voltage 


FIG.  395. — Graphical  method  of  finding  current  in  AB,  Fig.  393. 

across  generator  can  be  found  by  means  of  vector  diagrams 
in  Fig.  396.     Voltage  E  of  generator  =  134  volts  02  =  17.5°. 

Power  factor  of  load  =cos  17.5°  =.95. 

Power  delivered  by  gene rator  =  134  X  58  X. 95  =7.400  K.W. 
Power  used  in   motor  =110X40X.85=3.74   K.W. 

Power  used  in  lamps  =117X20X1     =2.34   K.W. 

Total  useful  power.  =6.08  K.W. 

(\  HQ 

Efficiency  of  transmission  =^-rj-=S2.2%. 


117  Vofts 


18.5 
FIG.  396. — Graphical  method  of  finding  voltage  across  generator,  Fig.  393. 

It  is  clear  from  above  example  that  the  lamps,  having 
unity  power  factor,  raise  the  power  factor  of  the  load  and 
that  if  some  piece  of  electrical  machinery  having  capacity 
reactance  were  placed  in  parallel  with  the  motor  and  lamps, 
the  power  factor  could  be  raised  still  more. 

281.  Two-phase  Distribution.  In  the  distribution  of 
Direct  Current,  we  have  seen  that  two  generators  may  be 
used  in  series  and  the  power  be  distributed  by  three  wires. 


>06 


ELEMENTS  OF   ELECTRICITY 


This  allows  a  voltage  to  bo  obtained  across  the  outside  of 
the  two  generators  which  is  twice  that  between  either 
wire  and  the  neutral,  as  explained  in  Chapter  IX.  By  means 
of  a  special  armature,  one  generator  is  usually  made  to 
supply  current  to  this  three-wire  system,  though  the 
principle  is  the  same  as  in  the  two-generator  plan. 


FIG.  396o. — Revolving  field  of  alternator-coils  excited  from  separate  source  of 
direct  current. 

In  the  same  way  two  A.C.  generators  can  be  run  in  series 
and  made  to  supply  current  to  a  three-wire  system.  But  the 
voltage  across  the  outside  of  the  generator  system,  need  not 
necessarily  be  twice  that  between  an  outside  wire  and  the  neutral, 
depending  on  whether  or  not  the  voltages  of  the  two  genera- 
tors were  in  phase.  This  voltage,  however,  would  always 
be  the  vector  sum  of  the  voltages  of  the  seoarate  generators. 


ALTERNA  TING  C  URRENTS 


507 


In  the  case  of  A.C.  generators  as  in  D.C.,  the  armature 
is  wound  so  that  one  generator  delivers  the  power  to  the 
three  wires.  Each  leg  is  then  called  a  PHASE,  and  the  sys- 
tem is  called  a  TWO-PHASE  SYSTEM.  For  several  reasons, 
the  voltages  of  the  two  phases  are  not  in  phase,  but  at  an 
angle  of  90°  to  each  other. 


FIG.  3966. — Stationary  armature  of  alternator.    No  moving  wires  or  contact  points 
to  insulate  for  the  high  voltage  generated. 

Figs.  396a  and  b  show  the  usual  construction  of  alternat- 
ing current  generators.  The  field  revolves  while  the  arma- 
ture is  stationary.  In  using  high  voltages  better  insulation 
can  be  obtained  by  this  construction. 

The  construction  and  action  of  such  a  machine  can  easily 
be  understood  by  reference  to  the  following  diagrams. 


508 


ELEMENTS    OF    ELECTRICITY 


Figs.  310  to  316  inclusive,  show  the  action  of  a  single  coil, 
single-phase  generator.     Now  if  another  coil  be  added  at 
right  angles  to  the  coil  of  Figs.  310  to  315,  as  in  Fig.  397, 
the  actions  will  be  the  same  in  each 
coil,  with  the  exception  that  they 
will  take  place  90°  later  in  one  coil 
than  in  the  other.     Thus,  when  the 
voltage    across    coil  A  B    is    at    a 
maximum,  as  in  Fig.  397,  the  volt- 
age  across  A'B'   is   zero,  and  vice 
versa.       The     voltages     may     be 
represented,   then,  by  two    curves 
differing    in    phase    by  90°,  as    in 
Fig.     398.       The    vector    diagram 
would  be  that  of  Fig.  399,  in  which 
FiG.397.-simpie2-Phase2-Poie  E  equals   voltage   induced   in  AB, 
fShrother  Coilsareat90°to  and    El    the    voltage    induced    in 

A'B'. 

The  internal  armature  connections  and  the  connections 
to  the  receiving  circuit  determine  the  amount  of  the  voltage 


-4 


FIG.  398,-^-Sine  curves  of  the  phase  relations  of  the  E.M.F.'s  in  the  phases  of  a 
2-phase  generator. 

delivered  to  the  line.     There  are  three  ways  of  making  these 
connections : 

(1)  The  ends  A  and  B  may  be  brought  out  to  one  pair 
of  collecting  rings,  and  A'  and  B'  may  be  brought  out  to 
another  pair.  In  that  case,  we  would  have  two  separate 
single-phase  lines,  in  which  the  voltages  would  be  identical 


ALTERNATING  CURRENTS 


509 


\ 


in  value  but  differing  in  phase  by  90°.  This  might  be 
called  a  four- wire,  two-phase  system. 

The  voltage-current  relations  in  each  phase  could  be 
treated  entirely  separately  as  single- 
phase  relations.  Diagrams  of  these 
conditions  are  shown  in  Figs.  400 
and  401,  the  latter  being  the  con- 
ventional representation.  Note  that 
the  phases  are  absolutely  distinct 
from  each  other. 

(2)  The  two  coils  may  be  joined 
in  series,  as  end  A'  of  one  coil  to 
end  B  of  the  other,  and  the 
terminals  A  and  Br  brought  out 

to  one  pair  of  collecting  rings.  The  induced  voltages 
in  the  two  coils  would  merely  be  added  vectorially 
before  being  brought  out,  and  the  machine  would  deliver 
a  simple  single-phase  voltage,  the  curve  of  which  would 
merely  be  the  sum  of  the  voltage  curves  of  each  coil. 


FIG.  399.— Vector  diagram  of 
the  E.M.F.'s  in  the  phases 
of  a  2-phase  generator. 


.'        [Phase 
AB 


FIG.  400. — Two-phase  system  using 
four  wires. 


Phase 
•  A'B' 


FIG.  401. — Conventional  representation  of 
a  2-phase  system  using  four  wires. 


(3)  By  far  the  most  common  way,  is  to  join  the  two 
coils  in  series  as  above  and  then  bring  out  a  lead  from  the 
juncture  to  a  third  collector  ring.  This  ring  then  acts  as  the 
neutral  in  a  three-wire  system,  of  which  each  phase  forms  a 
leg.  The  voltages  in  the  two  phases  differ  from  each  other 


510 


ELEMENTS  OF  ELECTRICITY 


by  90°.  Fig.  402  represents  this  case.  Assume  the  voltage 
across  phase  AB  =  voltage  across  A'B'  =  11Q  volts.  Then 
in  the  line,  the  voltage  across  XN  and  YN  =-110  volts. 
Across  XY,  however,  it  would  be  not  220  volts,  but  the 
vector  sum  of  110  volts  and  110  volts  at  right  angles  to  each 
other.  Thus  in  Fig.  403,  OX  =  voltage  across  NX  and 
OY  =  voltage  across  NY;  OAr=voltage  across  XY.  If 
OX  =  OY  =  11Q,  then  OW  =  155  volts. 

Note  that  line  ON  is  the  diagonal  of  a  square,  of  which 
OY  and  OX  are  sides. 


FIG.  402. — Diagram  of  a  3-wire  2-phase 
system. 


FIG.  403. — Graphical  method  of 
finding  the  voltage  across  the 
outer  wires  of  a  3-wire  2-phase 
system. 


The  length  of  the  diagonal  of  a  square  equals  \/2  times 
the  length  of  one  side.     Thus, 


This  is  generally  stated  : 

The  voltage  across  two  phases  in  a  two-phase  system 
equals  V2  times  the  voltage  across  one  phase  (the  phases 
being  at  90°  to  each  other). 

The  current  and  voltage  distribution  in  such  a  system  is 
found  by  the  same  method  as  that  used  in  a  B.C.  three-wire 
system.  Care  must  be  exercised,  however,  in  combining 
the  currents  or  voltages  in  the  different  sections  vectorially 
instead  of  algebraically. 

282.  Three-phase  Distribution.  Instead  of  using  an 
armature  with  two  coils  at  right  angles  to  each  other  as  in 
Fig.  397,  three  such  coils  might  be  used  at  an  angle  of  120° 


ALTKKNA TING  CURRENTS 


511 


0 


to  one  another  as  in  Fig.  404.  The  voltage  induced  in 
each  of  the  three  coils  would  then  differ  from  the  other  by 
an  angle  of  120°.  Fig.  405  shows 
the  voltage  curves  of  such  an  arrange- 
ment. Such  a  combination  might 
also  be  connected  internally  and 
leads  brought  out  to  collecting  rings 
in  three  ways: 

(1)  The  coils  may  all  be  connected 
in  series   and  the  two  ends  brought 
out  to  two  rings.     This  would    re- 
sult in  a  single-phase   generator,  as 
the  voltages  would  merely  be  added 

vectorially  within  the  armature  and     FlG-  404.— simple  s-phase 

generator. 

the   vector   sum  be  supplied  to  the 

rings,  and  distributed  to  a  single  pair  of  wires. 

(2)  The  three  corresponding  ends  (as  A,  A',  A")  may  be 
connected  to  a  common  terminal  and  this  be  brought  out 


FIG.  405. — Sine  curves  of  voltage  in  a  3-phase  generator. 

to  a  collecting  ring.  The  other  ends  (B,  Bf,  B")  would 
then  be  brought  out  separately  to  three  other  rings.  We 
would  then  have  a  three-phase  system  in  which  all  three 
phases  used  a  common  return.  This  is  commonly  called 
a  Y  connection. 

The  conventional  diagram  for  such  an  arrangement  is 
shown  in  Fig.  406.  The  voltage  from  any  outside  wire 
(X,  Y,  or  Z)  to  the  neutral  N  would  be  the  voltage  of  one 
phase. 


512 


ELEMENTS   OF  ELECTRICITY 


When  we  desire  to  find  the  voltage  across  any  two  phases, 
we  must  add  the  voltages  in  the  two  phases  vectorially. 

Now,  although  the  coils  are  said  to  be  at  an  angle  of  120° 
with  one  another  and  the  voltages  to  differ  in  phase  by  120°, 


FIG.  406. — Diagram  of  a  3-phasc  system;  Y-connectcd. 

it  will  be  noted  by  inspecting  Fig.  404  that  if  we  join  any 
two  adjacent  ends  together,  that  the  coils  thus  joined  are 
really  at  an  angle  of  60°  and  that  the  phase  difference 
between  the  voltages  is  really  60°. 

Thus  in  combining  the  voltage  across  phase  AB  with  the 
voltage   across  phase  A'B',  we  must   construct   a   vector 


Fia  407. — Graphical  method  of  finding  the  voltage  across  any  two  phases  of  s* 
3-phase  system. 

diagram  as  in  Fig.  407.  Because,  when  A  is  joined  to  A', 
this  diagram  will  represent  the  true  phase  relation  between 
the  voltages  in  the  two  coils,  as  is  seen  by  an  inspection  of 
Fig.  404.  The  vector  sum,  then,  of  AB  and  A'B'  is  the 


ALTERNATING  CURRENTS 


513 


line  AC.  If  AB=A'B'  =  110,  AC  by  measurement  =  191 
volts.  But  the  resultant  of  two  equal  forces  at  an  angle 
of  60°  to  each  other,  equals  \/3  times  the  value  of  either 
force. 

Thus  the  voltage  across  any  two  phases  of  a  F-connected 
generator  equals   V3  times  the  voltage  across  any  single 
phase.     The    current    and    voltage    relations    of    a   loaded 
three-phase   system   are  com- 
puted as  in  a  two-phase.    Care 
must  be  taken  to  note  correctly 
the  phase  relations    of  voltage 
and    current    in   the    different 
phases.       When    the     higher 
voltage  is    desired,    there    is 
need  of   but  three,  instead  of 
four,  wires  in  the  line. 

(3)  Instead  of  joining  like 
ends  of  the  three  phases 
together  as  in  (2),  we  may 
connect  unlike  ends,&s  A  to  5', 
A'  to  B",  and  A"  to  B. 
This  forms  a  closed  loop.  The 
junctures  of  the  three  coils 
are  now  brought  out  to  three 
collecting  rings.  This  is  com- 
monly called  a  J  (Delta)  connection.  Fig.  408  is  the  con- 
ventional diagram  for  this  arrangement. 

It  would  appear  at  first  sight  that  we  had  formed  a  short 
circuit  of  the  coils,  and  that  the  current  would  circulate 
around  through  the  armature  coils. 

But  here,  again,  particular  attention  must  be  paid  to  the 
phase  relations  between  the  voltages  across  the  coils.  By 
constructing  a  vector  diagram,  we  see  that  the  vector  sum 
of  the  voltages  in  any  two  coils  is  equal  and  opposite  to 
the  voltage  in  the  third.  The  resulting  voltage  tending 
to  cause  current  to  circulate  around  through  the  coils  is 


FIG.  408. — Diagram  of  a  3-phase 
system  J-connected. 


514 


ELEMENTS  OF  ELECTRICITY 


then  zero.  By  an  inspection  of  Fig.  404,  we  see  that  in 
joining  the  coils  together  in  this  way,  we  joined  adjacent 
ends.  The  coils  thus  joined  are  at  an  angle  of  120°  to  one 
another,  and  the  voltages  induced  in  the  coils  are  also  at  an 
angle  of  120°.  Thus,  in  the  vector  diagram  of  the  voltage 
relations  of  the  three  coils,  Fig.  409,  the  vectors  represent- 
ing the  voltages  are  at  an  angle  of  120°.  The  resultant  or 
vector  sum  of  any  two  voltages,  such  as  A'B'  and  A"B" , 
would  be  AR,  which  is  equal  and  opposite  to  voltage  A B. 
Therefore  the  voltage  tending  to  cause  a  current  to  flow 
around  through  the  three  coils  joined  in  A  is  zero.  The 


— $, 


FIG.  409. — Vector  diagram  of  voltage 
relations  across  the  paths  of  a 
3-phase  armature,  J-connected. 


FIG.  410. — Vector  diagram  of  current  in 
armature  paths  and  line  wires  of 
3-phase  system,  J-connected. 


voltage  induced  in  each  coil  is  accordingly  available  to 
force  a  current  through  any  outside  line  which  may  be 
joined  to  the  ends  of  the  coil,  as  the  three-phase  line,  XZ, 
ZY,  and  XY  of  Fig.  408.  The  voltage  across  each  phase 
of  the  line  is  the  voltage  across  each  path  of  the  armature. 

By  inspection  of  Fig.  408,  it  will  be  seen  that  each  line 
wire  of  the  three-wire  system  is  fed  by  two  coils  of  the 
armature.  It  will  also  be  seen  on  close  inspection  that  the 
currents  in  any  two  coils  feeding  one  line  wire  are  at  an  angle 
of  60°  to  each  other.  The  current  flowing  in  the  line  wire 
will  then  be  the  vector  sum  of  the  currents  in  these  two 
armature  coils. 

If  the  load  on  the  three  phases  is  balanced,  the  current 


ALTERNATING  CURRENTS 


515 


in  each  line  wire  will  be  the  same,  and  the  current  in  each, 
armature  path  will  be  the  same. 

The  current  in  the  line  would  then  be  the  vector  sum 
of  two  equal  currents  in  the  armature  at  an  angle  of  60° 
to  each  other.  We  have  seen  that  the  vector  sum  of  two 
equal  quantities  making  an  angle  of  60°  with  each  other 


FIG.  411. — Three-phase  generator.     May  be  used 
a  synchronous  motor. 


a  rotary  condenser,  or 


is  equal  to  \/3  times  one  of  the  quantities.     This  current 
relation  is  usually  stated : 

The  current  flowing  in  each  line  wire  of  a  balanced  three- 
phase  system  equals  the  V3  times  the  current  in  each  arma- 
ture path.  This  will  be  clear  from  the  vector  diagram, 
Fig.  410,  of  current  in  line  wire  X\.  This  wire  is  fed  by 
coils  AB  and  A"B" ,  as  per  Fig.  408. 


516  ELEMENTS  OF  ELECTRICITY 

Let  OB  represent  the  vector  for  the  current  in  phase 
AB.  Then  vector  OB"  of  the  current  in  A"B" ',  makes  an 
angle  of  60°  to  AB,  sincje  the  phase  of  the  current  in  A"B" 
is  seen  by  Fig.  404  to  be  60°  behind  the  phase  of  current 
in  AB.  OX  then  represents  the  current  in  the  line  wire 
X,  and  is  seen  to  be  V3  times  either  OB  or  OB". 

The  current  and  voltage  relations  throughout  a  loaded 
three-phase  A  system  are  found  in  accordance  with  the 
foregoing  principles.  Care  must  be  taken  at  all  times 
to  properly  represent  the  phase  relations  in  the  different 
branches.  Fig.  411  is  an  illustration  of  a  3-phase  genera- 
tor, using  no  neutral  wire.  This  machine  may  also  be 
run  as  a  motor,  or  as  a  rotary  condenser,  to  improve  the 
power  factor  of  the  line. 


ALTERNATING  CURRENTS  517 


SUMMARY  OF  CHAPTER  XV 

An  ALTERNATING  CURRENT  flows  back  and  forth 
through  a  circuit. 

A  CYCLE  consists  of  one  complete  flow  back  and  forth. 

The  FREQUENCY  is  the  number  of  cycles  completed  in 
one  second,  symbol,  f. 

A  cycle  is  divided  into  360  degrees ;%  any  instant  of  the 
cycle  being  rated  as  so  many  degrees  and  called  the  PHASE. 

An  ALTERNATING  E.M.F.  or  an  alternating  current  has 
four  values : 

(1)  The  instantaneous  value,  symbols  e  and  i. 

(2)  The  maximum  value,  symbols  E  and  I. 

(3)  The  average  value,  symbols  avE  and  avl. 

(4)  The  effective  value,  symbols  Ef  and  If. 

There  are  three  ways  of  representing  an  alternating  current : 

(1)  By  curve;  generally  the  sine  curve; 

(2)  Vector  diagram; 

(3)  Algebraic  equation. 

All  alternating  currents  considered  in  this  chapter  are 
assumed  to  follow  the  sine  curve. 

Any  INSTANTANEOUS  VALUE  depends  upon  the  phase 
and  is  found  by  the  equations: 

i=I  sin  (f>, 
e  =  E  sin  <£,. 

where  0  =  angle  of  phase. 

The  AVERAGE  VALUE  of  alternating  currents  and  voltages 
equals  .636  times  the  maximum  value. 

The  EFFECTIVE  VALUE,  which  is  the  value  always 
meant  when  the  terms  voltage  and  current  are  used  alone, 
is  that  value  in  amperes  of  an  alternating  current  which  will 
produce  the  same  heating  effect  as  a  direct  current  of  the  same 
number  of  amperes.  The  effective  value  equals  the  square 
root  of  the  average  of  the  squares  of  all  the  instantaneous 
values,  and  is  equal  to  .707  times  the  maximum  value. 

The  current  may  either  be  IN  PHASE  with  the  voltage, 
LAG  BEHIND  the  voltage,  or  LEAD  the  voltage.  The  angle 
of  lead  or  lag  is  represented  by  the  letter  0. 

POWER.   When  the  current  is  in  PHASE  with  the  voltage, 

i=I  sin  0, 
P=IfEf 


518  ELEMENTS   OF   ELECTRICITY 

When  the  current  LAGS  BEHIND  the  voltage, 

i=I  sin  («/>  —  0), 
P  =  IfEf  cos  0, 

(cos  6  is  called  the  power  factor). 
When  the  CURRENT  leads  the  voltage, 

i=Isin  (0  +  0), 
P=  IfEf  cos  0. 

The  cause  of  any  lag  in  an  alternating  current  is  the  induc- 
tive reactance  of  the  circuit. 

INDUCTIVE  REACTANCE  is  the  opposition  offered  by 
self  induction  to  the  flow  of  an  alternating  current,  and 
is  measured  in  ohms  ;  symbol,  XL  .  XL  =  2?rf  L. 

The  cause  of  any  lead  in  an  alternating  current  is  the 
capacity  reactance  of  the  circuit. 

CAPACITY  REACTANCE  is  the  opposition  offered  by 
capacity  to  the  flow  of  an  alternating  current,  and  is  meas- 

ured in  ohms;  symbol,  X<..  Xc  = 


REACTANCE  is  any  combination  of  inductive  and  capac- 
ity reactance  symbol,  X. 

X=XL-Xc. 

With    inductance   alone   in   the   circuit   the   current   lags 
90  degrees;  with  capacity  alone,  the  current  leads  90  degrees. 

Voltage 

Current  =—         -, 
Reactance 

but  the  current  is  always  90  degrees  out  of  phase  with  the 
voltage. 

IMPEDANCE,  symbol  Z,  is  the  combination  of  reactance 
and  resistance,  and  is  found  from  the  following  equation: 

Impedance2  =  reactance2  +  resistance2, 
or 


Voltage  to  force  a  current  of  If  amperes  through  an  impedance 
of  Z  ohms  equals  the  product  of  current  times  impedance,  i.e., 


or 

E=IZ. 


ALTERNATING  CURRENTS  519 

The  angle  of  lead  or  lag,  in  the  case  of  a  circuit  containing 
impedance,  can  be  found  from  the  equation, 


Voltage  and  current  values  in  parallel  and  series  circuits 
are  found  by  adding  vector  quantities.  See  summation 
on  Page  (491). 

Alternating  current  is  at  present  distributed  either  by 
SINGLE  PHASE,  TWO  PHASE  or  THREE  PHASE  SYSTEMS. 

In  a  TWO-PHASE  SYSTEM  the  voltage  across  one  phase 
is  at  an  angle  of  90  degrees  to  the  voltage  across  the  other. 

In  a  THREE-WIRE  TWO-PHASE  system,  the  voltage 
across  the  two  phases  equals  V2  times  the  voltage  across 
one  phase. 

In  a  THREE-PHASE  SYSTEM  the  voltage  across  each 
phase  is  at  an  angle  of  120  degrees  to  the  voltage  across  the 
other  phases. 

When  Y  CONNECTED,  ^AND  BALANCED,  the  voltage 
across  any  phase  equals  \/3  times  the  voltage  between  one 
line  wire  and  the  neutral.  Neutral  need  not  be  used. 

The  current  in  each  path  in  the  armature  is  the  same  as 
the  current  in  one  phase. 

When  J  CONNECTED  and  balanced,  the  voltage  across 
any  phase  equals  the  voltage  across  any  one  path  in  the 
armature. 

The  current  in  each  line-  wire  equals  v'3  times  current  in 
each  armature  path. 


520  ELEMENTS  OF  ELECTRICITY 


PROBLEMS   ON   CHAPTER  XV 

[TFave  forms   of  voltage  and    current  in   the  following    problems  are 
assumed  to  be  sine  curves.] 

92-16.  What  is  the  instantaneous  value  of  an  alternating 
E.M.F.  when  it  has  completed  40°  of  its  cycle?  Maximum  value 
=  135  volts. 

93-15.  What  is  the  average  value  of  an  alternating  current 
which  has  an  instantaneous  value  of  14  amperes  at  the  35°  phase? 

94-15.  What  would  be  the  effective  value  of  current  in  Prob- 
lem 93? 

95-15.  If  current  of  Problem  93  is  caused  by  voltage  of  Problem 
92,  and  current  phase  is  35°  when  voltage  phase  is  40°,  compute: 
(a)  Angle  of  lead  or  lag;  (6)  Power  factor;  (c)  Instantaneous 
power;  (d)  Average  power. 

96-15.  The  equation  for  an  alternating  E.M.F.  is  e  =  152  sin 
0.  (a)  What  is  e,  when  0  =  70°?  (b)  When  0  =  135°?  (c) 
When  0  =  960°? 

97-15.  The  instantaneous  value  of  a  current  at  25°  phase  is 
80 amps.  What  is  the  equation  for  the  current? 

98-15.  The  equation  for  an  alternating  current  is,  ?'=141  sin 
(0—20°).  (a)  What  is  the  instantaneous  value  of  the  current 
when  0  =  70°?  (b)  When  0  =  240°? 

99-15.  What  are  the  instantaneous  values  of  the  voltage  when 
the  phase  values  are  (a)  35°,  (b)  95°,  (c)  520°.  The  equation  for 
the  voltage  is  e  =  280  sin  0. 

100-15.  Two  voltages  E  and  E^  are  impressed  upon  a  circuit 
in  series.  El  =  l40  volts  and  lags  35°  behind  E,  which  equals 
201  volts,  (a)  What  is  the  voltage  across  the  circuit  when  E 
is  at  its  80°  phase?  (6)  When  E  is  at  10°  phase?  (c)  When  E 
is  at  its  125°  phase? 

101-15.  Two  alternating  currents  /  and  1^  are  flowing  in  parallel 
branches  of  a  circuit.  7  =  42  amperes,  /i  =  20  amperes  and  lags 
35°  behind  /.  (a)  What  is  the  resultant  of  these  two  currents? 
(b)  What  is  the  phase  relation  between  the  resultant  current 
and/? 


ALTERNATING  CURRENTS  521 

102-15.  If  the  voltage  across  the  parallel  circuits  in  Problem 
101  is  110  volts  and  is  in  phase  with  the  resultant  current,  find: 
(a)  Power  in  branch  containing  /. 
(6)  Power  in  branch  containing  /j. 
(c)  Total  power  in  parallel  circuit. 

103-15.  Find  instantaneous  power  in  each  branch  of  circuit 
of  Problem  101,  when  /  is  at  the  60°  phase. 

104-15.  How  many  volts  are  necessary  to  force  25  amperes 
alternating  current  through  8  ohms  resistance? 

105-15.  (a)  How  many  watts  are  consumed  in  resistance  of 
Problem  104?  (6)  How  much  direct  current  would  be  necessary 
to  cause  same  heating  effect  as  this  alternating  current? 

106-15.  If  a  coil  of  8  ohms  inductive  reactance  and  of  neglible 
resistance  is  used  instead  of  the  resistance  of  Problem  104,  how 
many  volts  are  necessary  to  force  25  amperes  through  it? 

107-15.  What  would  be  the  equations  for  the  instantaneous 
values  of  voltage  and  current  in  Problems  104  and  106? 

108-15.  (a)  If  the  coil  in  Problem  106  were  capacity  reactance, 
what  voltage  would  be  needed  to  force  25  amperes  through  it? 
(b)  Give  equations  for  instantaneous  values  of  current  and  voltage. 

109-15.  A  coil  of  .2  henry  inductance,  negligible  resistance, 
is  placed  on  a  110  volt  60  cycle  circuit.  What  current  flows? 

110-15.  A  coil  of  .2  henry  inductance  and  15  ohms  resistance, 
is  placed  in  a  110  volt  60  cycle  line,  (a)  What  current  flows? 
(b)  What  is  the  angle  of  phase  difference  between  current  and 
voltage  ? 

111-15.  (a)  Write  the  equations  for  the  instantaneous  values 
of  voltage  and  current  in  coil  of  Problem  110.  (b)  How  much 
power  is  consumed  by  the  coil? 

112-15.  Write  the  equations  for  the  instantaneous  values 
of  voltage  and  current  in  coil  of  Problem  109.  (b)  How  much 
power  is  consumed  by  the  coil? 

113-15.  If  the  frequency  in  Problem  110  were  changed  to 
25  cycles,  what  would  be  the  answers  to  Problems  110  and  111? 

114-15.  If  the  coil  in  Problem  110  were  put  on  a  110  volt  D.C. 
line,  what  would  be  the  answers  to  Problems  110  and  111? 

115-15.  When  the  voltage  across  coil  in  Problem  110  is  passing 
through  zero  (growing),  what  value  will  the  current  have? 


522  ELEMENTS  OF  ELECTRICITY 

116-15.  An  inductance  coil  has  a  resistance  of  80  ohms  and  an 
inductance  of  .4  henry.  When  on  a  40  cycle  line,  the  equation 
for  the  current  through  it  is  i  =  5  sin  <f>.  (a)  What  is  the  equation 
for  the  voltage  across  the  coil? 

117-15.  (a)  What  value  will  voltage  have  in  Problem  116, 
when  the  current  is  passing  through  zero  (growing)?  (b)  What 
power  is  consumed  by  the  coil? 

118-15.  A  choke  coil  of  negligible  resistance  and  .3  henry 
inductance  is  placed  in  series  with  a  6  ohm  resistance  across  a 
25  cycle  line.  The  current  which  flows  through  the  combination 
is  12.5  amperes.  What  is  the  voltage  across  each  and  across  the 
combination? 

119-15.  The  two  pieces  in  Problem  118  were  placed  in  parallel 
across  a  circuit  of  the  same  frequency  and  25  amperes  flowed 
through  the  choke  coil,  (a)  How  much  current  flowed  through 
the  resistance  and  how  much  current  flowed  in  the  main  line, 
assuming  these  two  were  the  only  pieces  in  the  circuit? 

120-15.  What  voltage  is  required  to  force  .1  ampere  through 
coil  in  Problem  1-10,  if  frequency  is  25  cycles  per  sec.  and  resistance 
of  coil  is  240  ohms? 

121-15.  What  current  will  flow  through  coil  in  Problem  5-10 
when  220  volts,  60  cycles,  are  put  across  the  terminals?  Resistance 
equals  400  ohms. 

122-15.  What  would  be  the  line  "  drop  "  in  transmitting  2 
amperes  A.C.  over  line  in  Problem  10-10?  Frequency =60. 

123-15.  If  50,000  volts  were  used  in  Problem  122-15,  how  much 
power  would  be  transmitted?  Non-inductive  load. 

124-15.  What  would  be  the  power  factor  and  efficiency  of 
transmission  in  Problem  123-15? 

125-15.  If  20  amperes  flow  in  an  A.C.  circuit  when  the  pressure 
is  115  volts,  what  are  the  resistance,  reactance  and  impedance? 
Power  factor  is  .85 ;  current  lags. 

126-15.  A  non-inductive  resistance  of  12  ohms,  a  capacity  of 
250  microfarads,  and  an  inductance  of  .08  henry  are  connected  in 
series  across  a  60  cycle,  112  volt  line.  Find: 

(a)  Voltage  across  the  resistance. 

(fr)  Voltage  across  the  inductance. 

(c)  Voltage  across  the  capacity. 

(J)  Current  through  the  combination. 


ALTERNATING  CURRENTS  523 

127-15.  (a)  Find  the  power  factor  of  combination  in  Problem 
126.  (?>)  Find  power  consumed  by  each  part  of  the  combination. 

128-15.  Two  inductance  coils,  having  resistances  of  12  and  6 
ohms  respectively,  and  inductances  of  .2  and  .08  henry  respec- 
tively, are  connected  irv  series  across  a  25  cycle  110  volt  circuit. 
Find: 

(a)  Current  through  the  coils. 

(b)  Voltage  across  each  coil. 

(c)  Phase  relation  of  voltage  across  first  coil  to  voltage  across 
second  coil. 

129-15.  If  coils  in  Problem  128  are  connected  in  parallel  across 
the  same  circuit,  what  will  be  the  answers  to  (a),  (b)  and  (c)? 

130-15.  A  generator  is  to  deliver  80  amperes  at  110  volts 
to  supply  power  to  incandescent  lamps,  which  are  non-inductive. 
If  the  line  wires  have  .4  ohm  resistance  and  .2  ohm  reactance, 
what  must  brush  potential  of  the  generator  be? 

131-15.  (a)  What  voltage  is  lost  in  the  Ime  of  Problem  130? 
(6)  What  is  efficiency  of  transmission? 

132-15.  An  induction  coil  is  connected  across  a  60  cycle  220 
volt  line  and  takes  12  amperes.  If  it  consumes  1400  watts  on 
this  circuit  what  power  will  it  consume  on  a  25  cycle,  110  volt 
line? 

133-15.  A  25  cycle  220  volt  generator  delivers  50  amperes  to 
run  induction  motors.  The  power  factor  is  then  .80.  What 
capacity  must  be  put  in  parallel  with  the  motors  to  bring  the 
power  factor  up  to  .95? 

134-15.  A  choke  coil  takes  8  amperes  and  consumes  1200  watts 
when  connected  across  a  60  cycle  220  volt  line.  What  resistance 
must  be  inserted  in  series  with  choke  coil  so  that  when  placed 
across  a  25  cycle,  110  volt  line  it  may  take  the  same  current? 

135-15.  A  synchronous  motor  takes  a  leading  current  of  45 
amperes  when  the  fields  are  overexcited.  An  induction  motor 
takes  a  lagging  current  of  85  amperes.  Power  factor  of  synchron- 
ous motor  is  .90;  of  induction  motor  .80.  If  the  two  motors  are 
operated  in  parallel  on  a  110  volt  line,  what  current  does  generator 
supply? 

136-15.  What  is  power  factor  of  load  on  generator  in  Prob- 
lem 135? 


524  ELEMENTS  OF  ELECTK1C1TY 

137-15.  In  a  three  wire,  two  phase  A.O.  system,  each  leg  has 
a  pressure  of  110  volts  across  it.  If  the  system  is  balanced  and 
40  amperes  are  flowing  in  each  phase,  what  current  flows  in  com- 
mon return  wire? 

138-16.  The  load  in  Problem  137  is  non-inductive.  What 
current  would  flow  in  each  phase  if  common  return  wire  were 
broken? 

139-15.  A  three  phase  delta-connected  generator,  has  equal 
loads  of  50  K.W.  J-connected  in  each  phase.  Voltage  across 
each  phase  equals  100  volts.  Unity  power  factor,  (a)  What 
is  the  current  in  each  phase?  (b)  In  each  lead  wire?  (c)  In  each 
armature  circuit. 

140-15.  If  generator  of  Problem  139  were  Y-connected,  what 
would  be  the  voltage  across  each  phase  of  the  line? 


APPENDIX 


USEFUL  NUMBERS 

v  1  ii  a     circumference  Q  ... 

7r  =  3.1416  = — -p .  Surface  of  cyl.  = 

diameter 

,T2=  9.8696;     ^  =  .3183.  Volume  of  cyl.  = 

7T 

Area  of  circle  =  nr2  =  —  =  .7854d2.      Surface  of  sphere  = 

/rd3     47rr3 

Volume  ol  sphere  =  ——  =  —5 — . 
b         o 

METRIC-ENGLISH  EQUIVALENTS 

1  cm.      =     .39  in.  1  in.       =     2.54  cms. 

1  m.        =39.37  ins.  1  ft.        =   30.48  cms. 

1  m.        =  3.23  ft.  1  ft.        =     .305  m. 

1  km.      =     .6  mile  1  mile    =     1.60  km. 

1  gm.      =     .035  oz.  (avoir.)  1  oz.       =   28.35  gms. 
1  kgm.    =    2.204  Ibs.  (avoir.)         1  Ib.       =453.6  gms. 

1  sq.cm.=      .154  sq.in.  1  sq.in.  =     6.45  sq.cms. 

1  cu.cm.=      .061  cu.in.  1  cu.in.  =   16.39  cu.cms. 

UNITS  OF  FORCE,  WORK,  POWER,  ETC. 

1  dyne  =   .00102  gm. 

Ift.lb.  =  1. 356  X107  ergs 

1  joule  =  107  ergs. 

1  horse-power =33,000  ft.lbs./min. 

1  horse-power  =      550  ft  .Ibs. /sec. 

1  horse-power  =  7. 46  XlO9  ergs /sec. 

1  horse-power  =  746  watts. 

1  watt  =   .00134  horse-power. 

1  watt  =  107  ergs. /sec.  =  1  joule/sec. 

MECHANICAL  EQUIVALENTS  OF  HEAT 
1  gm.  of  water  heated  1°  C.  =  4.2  X 107  ergs. 
1  Ib.  of  water  heated  1°  C.  =  1400  ft.lbs. 
1  Ib.  of  water  heated  1°  F.  =  778  ft.lbs. 
The  combustion  of  1  Ib.  of  coal  produces  about  14,000 
B,T,U. 

525 


526  APPENDIX 

SIGNIFICANT  FIGURES 

(From  Jameson's  "Mechanics,"  Longmans,  Green  &  Co.) 

The  results  of  all  experimental  work  should  be  so  ex- 
pressed as  to  indicate  as  nearly  as  possible  the  degree  of 
precision  with  which  the  work  was  performed.  It  is  evi- 
dent that  all  numbers  that  are  obtained  as  the  result  of 
measurements  are  limited  in  precision  by  the  nature  of 
the  apparatus  employed,  by  the  care  used  by  the  observer, 
the  size  of  the  units,  etc.  In  this  respect,  then,  such  quan- 
tities are  quite  different  from  the  pure  numbers  of  absolute 
value  as  employed  in  ordinary  arithmetical  operations. 

The  student  must  observe  carefully  the  following  rules 
in  all  laboratory  work  and  in  reports. 

I.  RECORDING  READINGS 

In  general,  scales,  etc.,  are  to  be  read  to  tenths  of  the 
smallest  divisions  marked  on  the  instrument.  The  last 
figure  entered  in  the  record  is  thus  assumed  always  to  be 
an  estimation  and  therefore  doubtful. 

Example  1. — 15.57  cms.  means  that  a  distance  was  meas- 
ured by  a  scale  subdivided  to  millimeters,  and  that  the 
observer  estimated  the  seven;  thus  the  distance  is  known 
to  be  between  15.5  and  15.6,  and  estimated  to  be  ^ths 
of  the  way  between  these  two  values.  It  is  misleading, 
and  furnishes  only  a  clue  to  what  we  actually  know  about 
this  distance  to  record  it  as  15.6  or  15.570  cm. 

Example  2. — A  distance  is  being  measured  with  a  rule 
subdivided  to  tenths  of  inches.  The  observer  finds  the  dis- 
tance to  be  as  nearly  exactly  seven  inches  as  he  can  dis- 
tinguish. This  should  be  recorded  7.00  in.  (not  7.0  or  7 
inb.).  Why? 

Example  3. — A  balance  is  capable  of  weighing  an  object 
to  .01  gm.  and  .001  gm.  can  be  estimated.  Notice  the 
correct  records  for  following: 


APPENDIX  527 

Eight  gms  ................    8.000  gins. 

Eight  and  ^  gms  ...........    8.500  gms. 

Eight  and  T^  -5-  gms  .........    8.070  gms. 

Eight  and  r^  gms  ........    8.008  gms. 

Eight-tenths  gm  ............  800  gms. 

In  general  a  series  of  readings  made  with  the  same  in- 
strument should  all  show  the  same  number  of  places  filled 
in  to  the  right  of  the  decimal  point  even  if  one  or  all  these 
places  are  zeros.  Why? 

It  is  often  convenient  to  express  in  decimal  form  read- 
ings taken  from  scales  divided  into  halves  of  units,  quar- 
ters, eighths,  etc.  In  all  such  cases,  retain  only  as  many 
places  in  the  decimal  as  correspond  approximately  to  the 
same  degree  of  precision  as  would  be  expressed  by  the 
fraction,  i.e.,  to  the  nearest  half  unit,  to  the  nearest  quar- 
ter, etc.  If  the  first  decimal  figure  rejected  is  5  or  greater, 
call  the  preceding  figure  one  larger  than  before. 

A  study  of  the  following  table  should  make  this  clear; 

i  =  .5  t   =.1  i  =  .4  §   =.7 

J  =  -3  A  =.06  f  =  .6  i   =.8 

J  =  .3  A  =.03  f  =  .9  A  -.19 

=  .2  =  -02  Etc-  Etc- 


II.  USE  OF  DATA  IN  CALCULATIONS 

Wherever  the  figure  following  the  doubtful  (last  re- 
tained) figure  is  5  or  greater  than  5,  increase  the  doubtful 
figure  by  unity.  Thus,  if  but  three  figures  are  to  be  kept, 
15.75,  15.76,  15.77,  15.78,  and  15.79  would  all  be  entered 
15.8. 

Notice  especially  that  the  location  of  the  decimal  point 
has  nothing  to  do  with  significant  figures.  Thus,  275,  27.5, 


528  APPENDIX 

2.75,  .275,  .0275,  .00275,  etc.,  are  all  results  expressed  to 
some  degree  of  precision,  and  in  each  there  are  three  and 
only  three  significant  figures,  the  5  being  the  doubtful  figure 
in  each. 

Averages. — In  averaging  a  series  of  determinations,  in 
general,  retain  in  the  result  the  same  number  of  significant 
figures  as  in  any  one  item. 

But  if  a  large  number  of  items  closely  agreeing  with  each 
other  are  averaged,  the  result  may  contain  one  more  sig- 
nificant figure  than  any  item. 

Multiplication. — After  the  operation,  keep  in  the  result 
as  many  figures,  counting  from  the  left,  as  there  are  sig- 
nificant figures  in  the  factor  having  the  lesser  number  of 
significant  figures. 

Division. — In  dividing  one  number  by  another,  keep 
in  the  quotient  as  many  figures  as  there  are  significant 
figures  in  the  number  having  the  lesser  number  of  signifi- 
cant figures.  Continue  the  divisions  only  far  enough  to 
determine  the  required  figures. 

Note  on  Multiplication  and  Division. — Ciphers  immediately  following 
the  decimal  point,  when  there  are  no  figures  to  the  left  of  the  point,  do 
not  count  as  significant.  Study  the  folio  whig  examples: 

(a)   15.75     X3.08        =48.5. 
(6)        .096   X   .096     =     .0092. 

(c)  .1523X   .00113=     .000172. 

(d)  720         X3.1          =2200. 

(e)  900         X800         =720000. 

In  (e)  only  the  first  cipher  is  significant.  It  is  necessary 
to  add  the  other  three  to  express  the  number  properly. 

(/)    325.6-^72.5  =  4.49. 

(g)         .0007859  -s- 157  =  .00000500. 


APPENDIX 


529 


Use  of  Pure  Numbers,  Constants,  etc. — In  using  pure 
numbers  and  constants  such  as  (3.1416),  .7854,  etc.,  do 
not  employ  more  figures  than  there  are  significant  figures 
in  the  values  depending  on  experiment  which  are  used 
with  them  in  the  same  calculation.  Thus  if  the  diameter 
of  a  circle-  is  measured  as  4.51,  the  area  is  4.51X4.51X 
.785  =  15.9.  The  use  of  more  numbers  in  the  constant 
lengthens  the  computation  and  gives  no  better  result. 
Why? 


CURVES 

Coordinate  Axes. — The  position  of  a  point  in  space  may 
be  fixed  by  reference  to  two   known  straight  lines  inter- 
secting  at  right   angles   in 
the  same  plane  as  the  point 
(OX  and  OF  of   the   Fig. 
412).     Such  lines  are  known 
as  coordinate  axes. 

The  horizontal  line  (OX) 
is  known  as  the  "  axis  of 
abscisses "  or  "  X  axis," 
the  vertical  line  (OF)  as 
the  "  axis  of  ordinates  "  or 


"  Y  axis."      The   point   of 

intersection  0  is  called  the  FIG  412. 

origin.      The  abscissa  of  a 

point  is  its  horizontal  distance  from  OF;    its  ordinate  is 

its  vertical  distance  from  OX.      These  given,  the  position 

of  the  point  is  determined.     Thus  P  is  that  point   which 

has  an  abscissa  of  3,  an  ordinate  of  5,  PI  the  point  which 

has  abscissa  of  11,  ordinate  of  8,  etc. 

For   convenience,   squared   or   "  cross-section  "   paper  is 
used  for  work  of  this  kind 


530 


Curves. — A  succession  of  related  points  may  be  connected 
by  a  smooth  line,  thus  constituting  a  "  curve."  Such 
curves  are  frequently  the  most  convenient  and  the  clear- 
est way  of  representing  a  physical  law,  corrections  for 
errors  of  apparatus,  etc.  Suppose,  for  example,  that  it  is 
desired  to  show  the  relation  between  the  stretch  of  a  wire 
and  the  stretching  loads  producing  it,  data  being  as  follows: 

Load.  Increase  in  Length. 

51bs 010  inch 

10   "    019     " 

15   "    030     " 

20    "    040     " 

25   "    051     " 

Taking  the  stretching  loads,  expressed  in  some  con- 
venient scale  of  lengths,  as  ordinates,  and  the  correspond- 
ing elongations  similarly 
expressed,  as  abscissae,  a 
series  of  points  may  be 
located  as  just  explained, 
and  through  these  a  smooth 
line  may  be  drawn.  In- 
spection of  the  curve  thus 
produced  (Fig.  413)  will 
show  at  a  glance  what 
could  be  obtained  from  the 
figures  only  on  more  ex- 
tended analysis.  The  law, 
"  Elongation  is  propor- 
tional to  the  load  applied,"  is  seen  immediately,  from  the 
nature  of  the  curve. 

Had  the  curve  turned  continuously  more  and  more 
toward  either  the  X  or  the  Y  axis,  showing  in  one  case  a 
progressive  increase,  in  the  other  a  progressive  decrease  in 


/ 

/ 

/ 

/ 

/ 

/ 

^/ 

.          .020             .040             .000             .08 
Elongation  in  Inches 
FIG.  413. 

APPENDIX  531 

elongation  with  increase  of  load,  or  had,  at  any  time,  a 
sudden  change  from  the  conditions  which  had  previously 
existed  occurred,  these  factors  would  have  been  brought  to 
the  attention  as  quickly. 

When  also,  as  here,  the  great  majority  of  points  lie  along 
a  straight  line  (or,  as  in  some  cases,  along  a  smooth  curve), 
any  experimental  errors  of  measurement  (as  in  the  elonga- 
tions for  loads  of  10  and  25  Ibs.),  will  be  shown  at  once  by 
the  fact  that  these  points  lie  slightly  off  the  line.  In  all 
such  cases,  the  curve  should  be  drawn  as  nearly  as  may  be 
through  all  points,  and  leaving  as  many  points  on  one  side 
as  on  the  other. 

The  student  must  in  all  cases  use  his  judgment  in  draw- 
ing the  curve  and  consider  the  conditions  of  the  experi- 
ment and  the  general  physical  law  illustrated. 

It  is  not  necessary,  and  indeed  often  not  advisable,  that 
ordinates  and  abscissae  be  expressed  in  the  same  scale.  Of 
course,  for  the  same  curve  all  abscissas  must  be  in  one 
scale,  and  all  ordinates  in  one  scale.  In  general,  the  scale 
adopted  should  be  that  most  convenient  for  the  particular 
values  which  will  at  the  same  time  give  a  curve  as  large 
as  the  paper  will  permit. 

One,  two,  five,  or  ten  units  to  a  square  will  be  found  the 
best.  Avoid  the  use  of  three  or  seven  units  per  square, 
or  other  inconvenient  subdivisions. 


GENERAL  DIRECTIONS  FOR  CURVE  SHEETS 

(1)  The  curve  must  be  done  neatly  in  India  ink. 

(2)  Heavy  lines  one  inch   in  from  the   margin  on  the 
ruled  portion  are  to  be  taken  as  axes,  except  where  all  the 
paper  is  necessary  for  the  curve.     The  origin,  i.e.,  the  in- 
tersection of  vertical  and  horizontal  axes,  should  be  at  the 
lower   left-hand    corner.     The   paper   may   be    used    with 


532  APPENDIX 

either  longer  or  shorter  side  as  vertical  axis,  according  to 
needs  of  the  curve. 

(3)  The   scale  on    which    the    curve    is    plotted    should 
be  so   selected    as   to  make   the  curve    as   large   as  pos- 
sible. 

(4)  Each  axis  should  be  marked  with  the  quantity  which 
it  represents,  and  with  the  unit  in  which  these  quantities 
are  expressed,   e.g.,   "  loads  in  pounds  per  square  inch," 
"  elongations  in  inches,"  etc.     These  titles  should  be  let- 
tered upon  the  ruled  paper  between  margin  and  axes. 

(5)  Eacn  half-inch  line  along  both  vertical  and  horizontal 
axes  should  be  marked  with  the  value  which  it  represents. 
No  other  figures  are  to  be  used  in  locating  the  curve. 

(6)  The  points  fixing  the  curve  are  to  be  located  by  a 
small  dot  around  which  is  drawn  a  small  circle  with  a  pair 
of  dividers. 

(7)  The  curve  should  usually  be  a  smooth  line  drawn  as 
nearly  as  possible  through  all  points.     It  will  represent  the 
most  probable  value  of  the  observations,  and  any  single 
point  lying  at  a  distance  on  either  side  of  the  line  will 
usually  be   a   result   of  error  in   observations.     Of  course 
judgment  must  be  used  in  drawing  this  conclusion,   and 
the  conditions  of  the  experiment  and  the  nature  of  the 
related  quantities  of  the  curve  must  always  be  taken  into 
account. 

(8)  The  name  of  the  student  and  the  date  should  be 
placed  at  the  bottom  of  the  sheet,  at  the  right,  in  small 
letters. 

(9)  The  title  of  the  curve  should  be  stated  in  the  lower 
right-hand  portion  of  the  curve  sheet  unless  this  interferes 
with  the  curve;    in  which  case  the  lower  left-hand  or  the 
upper  right-hand  portion  should  be  used. 

(10)  If  more  than  one  curve  is  drawn  on  the  same  paper 
for  comparison,  etc.,  use  the  same  origin  and  scales  for  alL 


APPENDIX 


533 


Distinguish  the  curves  by  the  title  printed  along  the  curve, 
or  by  lines  of  different  colors. 

(11)  All  titles,  explanations,  etc.,  must  be  in  lettering, 
and  no  handwriting  should  appear  upon  the  curve  sheet. 

THE  EQUATION  OF  A  STRAIGHT  LINE 

It  is  often  desired  to  find  the  equation  that  corresponds 
to  a  given  line  (straight  or  curved)  plotted  on  squared 
paper.  In  this  course  it  will  not  be  necessary  to  obtain 
the  equation  of  a  curved  line.  A  simple  method  for  the 
equation  of  a  straight  line  follows: 

Let  AB,  Fig.  414,  be  a  line  plotted  as  usual  on  the  axes 
OX  and  OY,  and  meeting  the  axis  of  Y  at  the  point  A. 
(If  the  line  as  first  drawn      Y 
does   not    cut   the   axis  of 
Y  it  must  be  extended  till 
it  does  so.) 

At  the  point  A  draw  a 
line  parallel  to  the  axis  of 
X.  Choose  any  point  on 


Fia.  414. 


the  line  as  P2,  and  draw 
its  ordinate  y2.     $2  is  the 
abscissa  of  this   point.     We  desire  to  obtain  an  equation 
that  will  give  us  the  relation  between  the  abscissa  and 
the  ordinate  for  this  and  every  other  point  on  this  line. 
We  notice  first  that  the  ordinate  y2  equals  the  intercept 
OA  on  the  Y  axis,  plus  P2D2,  or 


Also, 


and  so  on  for  every  point  on  the  line. 


534  APPENDIX 

The  value  of  the  intercept  OA  may  now  be  read  from 
the  curve.  Suppose  in  the  given  case  OA=S.  Next  read 
from  the  curve  values  of  the  altitude  and  base  of  any  tri- 
angle whose  hypothenuse  is  some  part  of  the  line  AB.  These 
values  are  to  be  expressed  in  units  of  the  respective  scales 
used  in  plotting  X  and  Y  and  not  as  actual  lengths  in 
inches.  The  triangle  AP2D2  will  serve.  Suppose  P2D2  =  4 

P2D2 

and  AD2  =  IQ  in  the  given   case.     Then  =.4.     But 

AJJ  '2 

AD2  =  x2,  therefore  P2D2  =  Ax2. 

If  we  had  used  other  triangles  we  should  have  obtained 
the  same  ratio  between  altitude  and  base,  and  thus, 


Or,  in  words,  we  may  now  say  that  any  ordinate  equals 
the  intercept  on  the  Y  axis  plus  .4  of  the  abscissa  for  the 
same  point.  Let  x  and  y  be  the  coordinates  of  any  point 
on  the  line  AB',  then 


which  is  the  equation  desired. 

P  D 

The  ratio  -—  —  is  sometimes  called  the  slope  of  the  line. 
AD2 

We  may  now  state  the  general  rule  as  follows: 
RULE.  —  The  equation  of  a  straight  line  is  formed  by 
putting  y  equal  to  the  intercept   on   the   axis  of   Y  plus 
the  slope  times  x.     If  intercept=a,  and  slope  (ratio)  =  ra, 
we  have, 

y=a-\-  mx. 


APPEXDlX  535 

NOTE. — The  student  will  notice  that  the  equation  just 
given  is  perfectly  general.  If  the  line  cuts  the  axis  of  Y 
below  the  origin,  the  intercept  will  be  a  negative  term  and 
the  equation  will  be  of  the  form  y=  —a  +  mx.  If  the  line 
slopes  so  that  an  increase  in  the  value  cf  the  abscissa  causes 
a  decrease  in  the  value  of  the  ordinate,  then  m  will  be  a 
negative  quantity,  y  =  a  —  mx.  It  is  possible,  of  course, 
that  both  a  and  m  may  be  negative  at  the  same  time,  as 
y  =  —  a  —  mx.  The  student  should  draw  and  consider  care- 
fully lines  to  illustrate  each  case. 

EXERCISES 
1.  Locate  following  points: 


Point, 
(a)    

Abscissa. 
5   

Ordinate. 
.     3 

(6)    

7  

10 

(c)    

5   

8 

(d)   

0   

12 

(e)    

5   

5 

m 

.   9   . 

0 

2.  Measure  carefully  the  lengths  of  9  ins.,  7  ins.,  5  ins., 
3  ins.,  respectively  in  metric  units.     Make  each  measure- 
ment three  times,  using  different  parts  of  the  scale.      Why? 
Take  the  average  of  the  three  readings  and  plot  a  curve, 
using  inches  as  ordinates  and  the  corresponding  number 
of  centimeters  as  abscissa. 

The  curve  will  pass  through  the  origin  or  zero-point  of 
each  scale.     Why? 

From  your  curve  find  the  value  of  1  inch  in  centimeters. 
What  is  the  true  value?     What  is  your  per  cent  of  error? 

3.  A  determination  of  the  relation  of  bending  of  beam  to 
load  gave  following  results: 


536  APPENDIX 

Loads.  Deflections. 

10  Ibs 0.05  inch. 

20   "    0.10     " 

30   "    0.15     " 

40   "    0.21     " 

50   "    0.25     " 

60   "    0.29     " 

70   "    0.35     " 

Plot  curve  showing  relation  of  deflection  to  load,  using 
loads  as  ordinates. 

4.  Plot  a  straight  line  such  that  it  shall  gain  3  units  of 
absciss®  for  every  unit  gained  as  ordinates. 

5.  A  determination  of  Boyle's   Law  gave  the  following 
data: 

Pressure.  Volume  of  Gas. 

82.1 12.03 

88.2 : 11.20 

96.2 10.26 

105.5 9.35 

118.9 8.31 

135.5 7.29 

160.1 6.17 

Plot  the  curve  of  above  values,  using  pressures  as  ordi- 
nates. 

6.  A  determination  showing  the  effect  of  length  of  a  beam 
upon  its  stiffness  gave  the  following  data: 

Length.  Deflection. 

10  inches 005  inch 

20   "   035  " 

30   "   120  " 

40   "  .285  " 


APPENDIX  537 

Plot    curve    showing    relation    of    deflections   to   length, 
using  deflections  as  abscissae. 

7.  The  area  of  a  circle  varies  with  square  of  its  diameter. 
Plot  a  curve  to  show  relation  of  area  to  diameter  in  circle 
whose  successive  diameters  are  1,  2,  3,  4,  5;    and  6.     Use 
diameters  as  ordinates. 

8.  What  is  the  equation  of  the  line  which  cuts  the  axis 
of  Y  at  a  point  3.4  above  the  origin  and  which  rises  5  units 
for  every  8  of  horizontal  distance? 

9.  If  the  intercept  of  a  line  is  .56  and  its  slope  is  2.58, 
what  is  its  equation? 

10.  A  line  has  two  points  whose   coordinates  are   (8.5) 
and  (3.1).     Plot  the  line  and  obtain  its  equation. 

11.  What  is  the  slope  of  a  line  if  its  intercept  is  4  and 
a:  =  12  when  #  =  6.8? 

,  12.  The  slope  of  a  certain  line  is  .532,  and  it  passes 
through  the  origin.  What  is  the  ordinate  of  a  point  on 
this  line  whose  abscissa  is  2.3? 

13.  Write  the  equation  of  the  following  lines,     a  =  inter- 
cept and  m  =  slope. 

(a)  a  =  5,  ra=     .13. 

(6)  a  =-5,  m=   3. 

(c)  a  =-2.3,  m=     .70. 

(d}  a  =  3,  ra=-.68. 

(e)  a=  -2.5,  w=-.45. 

14.  Draw  a  sketch  to  show  the  general  character  of  each 
of  the  lines  described  in  Problem  13. 


538  APPENDIX 

SIMPLE  TRIGONOMETRIC  FUNCTIONS 

Let  ABC,  Fig.  415,  be  any  given  angle,  and  let  P  be  any 
point  on  the  line  BC.  From  P  draw  PN  perpendicular  to 
the  side  BA,  thus  forming  the  right-angled  triangle  PNB. 
Suppose  we  measure  BP  and 
find  it  to  be  10  inches  long  and 
PN  and  find  it  to  be  5.05  inches 
long.  Then 


BP          10  FIG.  415. 

Now  let  us  choose  another  point  on  BC,  as  P',  draw  the 
perpendicular  P'N'  and  measure  lines  BP'  and  P'Nf.  If 
BP'  is  |  as  long  as  BP,  or  8  inches,  then  P'N'  will  be  | 

P'N' 

as  long  as  PN,  or  5.05X4=4.04  inches.     The  ratio  — —  = 

BP 

4.04 

—  =  .505  or  the  same  as  that  of  PN  to  BP.     So  we  might 
8 

choose  any  point  on  BC  and  always  obtain  the  same  ratio 
of  the  perpendicular  to  the  hypothenuse  so  long  as  we  use 
the  same  angle  ABC. 

BN 

Similarly,   — —   gives   a   second    constant   ratio   for    the 
BP 

PN 

given  angle,  and  "-— —  a  third,  all  of  which  are  independent 
BN 

of  the  position  of  P  and  dependent  only  on  the  angle  ABC. 

PN 

If  the  angle  ABC  be  changed,  then  the  ratios  -^^»   e*c-? 

nr 

will  have  new  values,  but  these  again  will  be  the  same, 
no  matter  where  P  is  taken  on  the  line  BC.  Each  angle 
thus  has  a  certain  number  of  constant  ratios  among  which 


APPENDIX  539 

are  the  three  here  given,  and  these  ratios  are  given  distin- 
guishing names. 

PN 

The  ratio  — —  is  called  the  sine  of  the  angle  ABC. 

DL 

BN 

The  ratio  — —  is  called  the  cosine  of  the  angle  ABC 

PN 

The  ratio  — —  is  called  the  tangent  of  the  angle  ABC. 
BN 

Definitions. — Let  B  be  an  angle  (not  the  right  angle)  of 
a  right-angled  triangle.  The  sine  of  the  angle  B  is  the 
ratio  of  the  side  opposite  the  angle  to  the  hypothenuse  of 
the  triangle. 

The  cosine  of  the  angle  B  is  the  ratio  of  the  side  adja- 
cent the  angle  to  the  hypothenuse  of  the  triangle. 

The  tangent  of  the  angle  B  is  the  ratio  of  the  side  oppo- 
site to  the  side  adjacent. 

In  calculations,  sine,  cosine,  and  tangent  are  always 
written  for  brevity,  sin,  cos,  tan.  Thus,  sin  30°  =  .500; 
cos  45°  =  .707;  tan  50°  =  1.19. 

The  values  of  these  ratios  have  been  calculated  for  all 
angles,  and  are  given  in  what  are  called  tables  of  trigono- 
metric functions.  Such  tables,  with  the  values  carried  out 
to  three  decimals,  will  be  found  on  the  following  page. 


540 


APPENDIX 


TRIGONOMETRIC  FUNCTIONS 


A 

Sin. 

Cos. 

Tan. 

A 

Sin. 

Cos. 

Tan. 

0 

.000 

1.000 

.000 

1 

.017 

.999 

.017 

46 

.719 

.695 

1.04 

2 

.035 

.999 

.035 

47 

.7IU 

.682 

1.07 

3 

.052 

.999 

.052 

48 

.743 

.669 

.11 

4 

.070 

.998 

.070 

49 

.755 

.656 

.15 

5 

.087 

.996 

.087 

50 

.766 

.643 

.19 

6 

.105 

.995 

.105 

51 

.777 

.629 

.23 

7 

.122 

.993 

.123 

52 

.788 

.616 

.28 

8 

.139 

.990 

.141 

53 

.799 

.602 

.33 

9 

.156 

.988 

.158 

54 

.809 

.588 

.38 

10 

.174 

.985 

.176 

55 

.819 

.574 

.43 

11 

.191 

.982 

.194 

56 

.829 

.559 

.48 

12 

.208 

.978 

.213 

57 

.839 

.545 

.54 

13 

.225 

.974 

.231 

58 

.848 

.530 

.60 

14 

.242 

.970 

.249 

59 

.857 

.515 

.66 

15 

.259 

.966 

.268 

60 

.866 

.500 

.73 

16 

.276 

.961 

.287 

61 

.875 

.485 

.80 

17 

.292 

.956 

.306 

62 

.883 

.469 

.88 

18 

.309 

.951 

.325 

63 

.891 

.454 

.96 

19 

.326 

.946 

.344 

64 

.898 

.438 

2.05 

20 

.342 

.940 

.364 

65 

.906 

.423 

2.14 

21 

.358 

.934 

.384 

66 

.914 

.407 

2.25 

22 

.375 

.927 

.404 

67 

.921 

.391 

2.36 

23 

.391 

.921 

.424 

68 

.927 

.375 

2.48 

24 

.407 

.914 

.445 

69 

.'934 

.358 

2.61 

25 

.423 

.906 

.466 

70 

.940 

.342 

2.75 

26 

.438 

.898 

.488 

71 

.946 

.326 

2.90 

27 

.454 

.891 

.510 

72 

.951 

.309 

3.08 

28 

.469 

.883 

.532 

73 

.956 

.292 

3.27 

29 

.485 

.875 

.554 

74 

.961 

.276 

3.49 

30 

.500 

.866 

.577 

75 

.966 

.259 

3.73 

31 

.515 

.857 

.601 

76 

.970 

.242 

4.01 

32 

.530 

.848 

.625 

77  - 

.974 

.225 

4.33 

33 

.545 

.839 

.649 

78 

.978 

.208 

4.70 

34 

.559 

.829 

.675 

79 

.982 

.191 

5.14 

35 

.574 

.819 

.700 

80 

.985 

.174 

5.67 

36 

.588 

.809 

.727 

81 

.988 

.156 

6.31 

37 

.602 

.799 

.754 

82 

.990 

.139 

7.12 

38 

.616 

.788 

.781 

83 

.993 

.122 

8.14 

39 

.629 

.777 

.810 

84 

.995 

.105 

9.51 

40 

.643 

.766 

.839 

85 

.996 

.087 

11.43 

41 

.656 

.755 

.869 

86 

.998 

.070 

14.30 

42 

.669 

.743 

.900 

87 

.999 

.052 

19.08 

43 

.682 

.731 

.933 

88 

.999 

.035 

28.64 

44 

.695 

.719 

.966 

89 

.999 

.017 

57.28 

45 

.707 

.707 

1.000 

90 

1.000 

.000 

Infinity. 

APPENDIX 


541 


TYPICAL  VALUES   OF   B  AND   H  FOR  DIFFERENT  IRONS 
(From  Caldwell's  "Electrical  Problems") 


B  in  Gausses. 

H  in  Gausses. 

Sheet  Steel. 

Cast  Steel. 

Wrought  Iron. 

Cast  Iron. 

3,000 

1.3 

2.8 

2.0 

5.0 

3,500 

1.4 

3.1 

2.2 

6.5 

4,000 

1.6 

3.4 

2.5 

8.5 

4,500 

1.7 

3.7 

2.7 

11.0 

5,000 

1.9 

3.9 

3.0 

14.5 

5,500 

2.1 

4.2 

3.2 

18.5 

6,000 

2.3 

4.5 

3.5 

24.0 

6,500 

2.4 

4.8 

3.7 

30.0 

7,000 

2.6 

5.1 

4.0 

38.5 

7,500 

2.8 

5.4 

4.2 

49.0 

8,000 

3.0 

5.8 

4.5 

60.0 

8,500 

3.2 

6.1 

4.7 

74 

9,000 

3.5 

6.5 

5.0 

89 

9,500 

3.7 

7.0 

5.3 

106 

10,000 

3.9 

7.5 

5.6 

124 

10,500 

4.1 

8.2 

6.0 

144 

11,000 

4.4 

9.0 

6.5 

166 

11,500 

4.7 

11.2 

7.2 

191 

12,000 

5.0 

11.5 

7.9 

222 

12,500 

5.5 

13.7 

8.9 

255 

13,000 

6.0 

16.0 

10.0 

290 

13,500 

7.0 

18.0 

13.0 

328 

14,000 

9.0 

21.5 

15.0 

369 

14,500 

12.0 

27.0 

18.5 

15,000 

15.5 

32.0 

25.0 

15,500 

20.0 

40.0 

35.0 

16,000 

27.0 

49.0 

49.0 

16,500 

37.5 

60.0 

69.0 

17,000 

52.5 

74.0 

93.0 

17,500 

70.0 

93.0 

120 

18,000 

92.0 

115 

152 

18,500 

119 

139 

186 

19,000 

149 

175 

229 

19,500 

189 

226 

277 

20,000 

232 

285 

20,500 

277 

21,000 

327 

21,500 

383 

22,000 

441 

22,500 

498 

23,000 

555 

23,500 

612 

24.000 

669 

24,500 

726 

25,000 

783 

542 


APPENDIX 


Material  (Commercial). 

Resistivity;  Ohms 
per  Mil-Foot 
at  20°  C 

Temperature  Coeffi- 
cient of  Resistance  = 
Increase  per  degree  C. 

Resistance  at  0° 

Aluminum  

17.4 

00435 

Copper  annealed 

10  4 

0042 

Coppsr  hard  drawn 

10  65 

Iron  annealed 

90 

005 

Iron  EBB.  (Roebling)   

64 

0046 

German  silver     

114  to  275 

00095 

Manjranin          

250  to  450 

00001 

I  \IA  (Boker)  soft 

283 

000005 

IAIA.  (Boker)  hard 

300 

00001 

Advance  (Driver-Harris)  

294 

.  00000 

RESISTANCE  OF  SOFT  OR  ANNEALED  COPPER  WIRE 


B.  &S. 
Gauge. 

Diameter 
in  Mils,  d. 

Area  in 
Circular 
Mils.  d2. 

Ohms 
Per 

1000  Ft. 
at  20°  C. 
or68°F. 

B.  &S. 
Gauge. 

Diameter 
in  Mils,  d. 

Area  in 
Circular 
Mils,  d*. 

Ohms 
Per 

1000  Ft. 
at  20°  C. 
or  68°  F. 

0000 

460.00 

211,600 

.04893 

21 

28.462 

810.10 

12.78 

000 

409.64 

167,810 

'.06170 

22 

25.347 

642.40 

16.12 

00 

364.80 

133,080 

.07780 

23 

22.571 

509.45 

20.32 

0 

324.86 

105,530 

.09811 

24 

20.100 

404.01 

25.63 

25 

17.900 

320.40 

32.31 

1 

289.30 

83,694 

.1237 

26 

15.940 

254.10 

40.75 

2 

257.63 

66,373 

.1560 

27 

14.195 

201.50 

51.38 

3 

229.42 

52,634 

.1967 

28 

12.641 

159.79 

64.79 

4 

204.31 

41,742 

.2480 

29 

11.257 

126.72 

81.70 

5 

181.94 

33,102 

.3128 

30 

10.025 

100.50 

103.0 

6 

162.02 

26,250 

.3944 

31 

8.928 

79.70 

129.9 

7 

144.28 

20,816 

.4973 

32 

7.950 

63.21 

163.8 

8 

129.49 

16,509 

.6271 

33 

7.080 

50.13 

206.6 

9 

114.43 

13,094 

.7908 

34 

6.305 

39.75 

260.5 

10 

101.89 

10,381 

.9972 

35 

5.615 

31.52 

328.4 

11 

90.742 

8,234.0 

1.257 

36 

5.000 

25.00 

414.2 

12 

80.808 

6,529.9 

1.586 

37 

4.453 

19.82 

522.2 

13 

71.961 

5,178.4 

1.999 

38 

3.965 

15.72 

658.5 

14 

64.084 

4,106.8 

2.521 

39 

3.531 

12.47 

830.4 

15 

57.068 

3,256.7 

3.179 

40 

3.145 

9.89 

1047 

16 

50.820 

2,582.9 

4.009 

17 

45.257 

2,048.2 

5.055 

18 

40.303 

1,624.3 

6.374 

19 

35.890 

1,288.1 

8.038 

20 

31.961 

1,021.5 

10.14 

APPENDIX  543 

TABLE  OF  CARRYING  CAPACITY  OF  WIRES 

(NATION  \L  ELECTRIC  CODE) 

The  following  table,  showing  the  allowable  carrying  capacity  of 
copper  wires  and  cables  of  98  per  cent  conductivity,  according  to  the 
standard  adopted  by  the  American  Institute  of  Electrical  Engineers, 
must  be  followed  in  placing  interior  conductors. 

(For  insulated  aluminum  wire,  the  safe  carrying  capacity  is  84  per 
cent  of  that  given  in  the  following  tables  for  copper  wire  with  the 
same  kind  of  insulation.) 


Table  A. 

Table  B. 

Table  A. 

Table  B. 

B.  &S. 
Gauge. 

Rubber 
Insulation, 

Other 
Insulations, 

Circular  Mils. 

Rubber 
Insulation, 

Other 
Insulations, 

Amperes. 

Amperes. 

Amperes 

Amperes. 

18 

3 

5 

200,000 

200 

300 

16 

6 

8 

300,000 

270 

400 

14 

12 

16 

400,000 

330 

500 

12 

17 

23 

500,000 

390 

590 

10 

24 

32 

600,000 

450 

680 

8 

33 

46 

700,000 

500 

760 

6 

46 

65 

800,000 

550 

840 

5 

54 

77 

900,000 

600 

920 

4 

65 

92 

1,000,000 

650 

,000 

3 

76 

110 

1,100,000 

690 

,080 

2 

90 

131 

1,200,000 

730 

.150 

1 

107 

156 

1,300,000 

770 

,220 

0 

127 

185 

1,400,000 

810 

,290 

00 

150 

220 

1,500,000 

850 

,360 

000 

177 

262 

1  ,600,000 

890 

,430 

0000 

210 

312 

:  1,700,000 

930 

,490 

1,800,000 

970 

,550 

1,900,000 

1,010 

,610 

2,000,000 

1,050 

1,670 

The  lower  limit  is  specified  for  rubber-covered  wires  to  prevent 
gradual  deterioration  of  the  high  insulations  by  the  heat  of  the  wires, 
but  not  from  fear  of  igniting  the  insulation.  The  question  of  drop 
is  not  taken  into  consideration  in  the  above  tables. 

The  carrying  capacity  for  No.  16  and  No.  18  B.  &  S.  gauge  wire  is 
given,  but  no  smaller  than  No.  14  is  to  be  used. 


544 


APPENDIX 


y. 

P 

I« 


1  British  Th 
Symbol,  B. 
253  calories. 


. 


^"73»ccg«Oi-< 

s:3a3S=2 


122  watt  per  pq.in. 
0176  K.W.  per  sq. 
0236  H.-P.  persq.: 


. 
0  ft.-lbs.  per  hour 
t.-lbs.  per  minute. 
.-lbs.  per  second. 
at-units  per  hour. 
t-units  per  minut 
t-unit  per  second. 
.  carbon  oxidize 

.  water  evap.  per 
nd  at  212°  F. 


2,654, 
240 
.3  f 
12 
h 
h 


1,00 
1.34 
2,65 
44,2 
737. 
3,41 
56.9 
.948 
.227 
hr. 
3.53 
fr 


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c  E  >-  i-  o>^3  ? 

bi&£fct5!i 

a^^sg.0^^ 

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EMg^J^Jl 

sgss^gss* 

I  N-^WiOW      ^t    .« 


3  3 

O  O 

JS  JC 

t-'       o  c 

3      .     v  a 

-•sifa 
-s  IJfeiS 

WlIJl!, 

WJ'^«3!t8 

g^jz*.-  g^  -a 

3  JC  IM  «  1C  C  "t        5- 

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d- 

o     *• 
a    c- 


PLATE  I 


FIG.  34. — Browning  electric  lift  magnet  in  action. 


MAGNET  FPAME-  SINGLE  POINT  SUSPENSION 


TWIN  CONDUCTOR 
TAUC  FLEXIBLE 
CONDUIT. 


\ 


MAGNET  Con. 
OUTER  POLE. 


-MAGNET  RELEASE  DIAPHRAM. 
INNER    POLE. 
NON-MAQ.NETIC  BOTTOM  PLATE. 


BROWNINQ  ELECTRIC 


INDEX 


Abscissa,  definition  of,  529 
Action  within  armature  of  genera- 
tor, 170 
of  motor,  205 
Ageing  of  magnets,  19 
Alternating  current,  430 

circuits,  general  law  for,  473 
parallel,  current  and  voltage 

relation  in,  180 
summary  of  relations  in,  491 
series,    combined     reactance 

in,  479 

combined  resistance  in,  479 
voltage  across,  475 
summary  of  relations  in,  491 
current,  average  value  of,  443 
effective  value  of,  444 
computation  of,  442 
curves,  adding  of,  482 
E.M.F.,  average  value  of,  441 
effective  value  of,  445 
instantaneous  value  of,  439 
hydraulic  analogy,  431 
inductive  reactance,  454 
lagging  current,  450 

curve  of,  454 
leading  current,  452 
phase  relation  of  current  and 

voltage,  446 
power,  492 

measurement  of,  497 
summary  of  current  and  voltage 

relation,  491 
vector  diagrams  of,  435 
wattless  component  of,  496 


Alternating     current      machines, 
generator,  armature  wind- 
ings, 513 
computation  of  brush  potential, 

501 

elementary  form,  161,  432 
rotating  armature,  515 
rotating  field,  506,  507 
three-phase,  510 
two-phase,  509 
motors,  induction,  496,  497 

power  factor  of.  497 
rotary  condenser,  515 
synchronous,  515 

power  factor  of  as  a  rotary 

condenser,  515 
Ammeter,  401 

connection  of,  in  circuit,  70 
electro-dynamometer  type,  404 
hot-wire  type,  403 
permanent  magnet,  406 
principle  of  operation  of  Weston 

type,  27 

resistance  of,  72,  401 
solenoidal  type,  401 
summary  of  types,  416 
Thomson     inclined     coil    type, 

403 

two-coil  type,  404 
Weston  type,  406 
Ampere,  definition  of,  37,  335 
Ampere-turns,  defirition  of,  132 

computation  of,  132 
Angle  of  lead,  on  generator,  174 

on  motor,  206 
Anode,  333 

547 


548 


INDEX 


Arc  lamps,   ballasting  resistance 

in,  375 
distribution  of  illumination  of, 

373 

operation  of  D.C.,  375 
regulating  coil  in,  376 
Armature  paths,  177 
Armature  reaction,  of  generator, 

170 

of  motor,  205 
Armature     resistance,     of     D.C. 

generator,  176 
Astatic  galvanometer,  388 
Average    value  of   A.C.    current, 

443 

Average  value  of  E.M.F.,441 
Axis  of  commutation,  of  genera- 
tor, 173 
of  motor,  205 

Axis  of  least  sparking,  of  genera- 
tor, 172 
of  motor,  206 
Ayrton  shunt,  393 

B 
B  (flux  density),  131,  140 

table  of  typical  values,  541 
Backward  lead  of  brushes,  206 
Balanced  three-wire  system,  242 
Ballasting  resistance  in  arc  lamps, 

375 
Ballistic    galvanometer,     use    in 

measuring  capacity,  311 
Battery,  324  (see  Storage  Battery) 

booster  used  with,  352 

capacity  of  lead  battery,  343 

care  of  storage  battery,  342 

chemical  action  of,  340 

discharge  rate,  343 

dry,  325 

efficiency  of  lead  battery,  337, 
338 

E.M.F.  of,  325 

floating  battery,  353 

internal  resistance  of,  347 

terminal  voltage  of,  337 

troubles,  342 
Bolometer,  398 


Booster,  351 

Bound  charges,  303 

Bridge,  Wheatstone,  107 

method  of  measuring  capacity, 
312 

Brush  potential,  50 

computation  of,  in  A.C'.  genera- 
tors, 501 

Building  up  of  generator  voltage, 
183 


Candle-foot,  see  Foot-candle 
Candle  power,  365 

measurement  of,  368 
Capacity,  300 

measurement  of ,-311 

of  aerial  twin  wires,  310 

of  plate  condensers,  308 

of  sheathed  cables,  309 

of     telephone     and     telegraph 
cables,  308 

unit  of,  300 
Capacity  reactance,  459 

computation  of,  465 
Carrying  capacity  of  wires,  543 
Carbon  filament,  377 
Cathode,  333 

Chloride  accumulator  storage  bat- 
tery, 339 

Circuit-breaker,  principle  of  opera- 
tion of,  32 
Circuit,  magnetic,  2 
Circular  mil,  96 
Closed  circuit  cell,  328 
Coercive  force,  148 
Collecting  rings,  165 
Combined  impedance  of  series  A.C. 

circuits,  479 

Commercial  efficiencj^  of  generator, 
257 

of  motor,  258 

Commutating  pole  motor,  212 
Commutator,  165 
Comparison  of  Edison  and   lead 
types  of  storage  battery,  357 

of  generator  and  motor  charac- 
teristics, 223 


INDEX 


Compass,  10 

action  of,  13 
Compensating  coil  of  wattmeter, 

418 
Compound  generator,  180 

motor,  221 
Condensers,  305 

in  parallel,  316 

in  series,  317 
Conductance,  56 
Consequent  poles,  17 
Constant  current  generator,  181 
Control,  definition  of,  210 

of  current  in  circuits,  95 

of  motor  speed,  210 
Cooper-Hewitt  lamp,  381 
Copper  loss,  in  D.C.  machines,  253 

in  generators,  188 
Core  magnetization,  in  armature 

of  generator,  170 
Core  type  transformer,  278 
Cosine,  defined,  539 
Coulomb,  37 
Counter  E.M.F.,  203 
Cumulative  compound  motor,  221 
Current,  electric,  nature  of,  37 

computation  of,  442 

effective  value  of,  444 

in  A.  C.  circuits,  average  value 
of,  443 

in  armature  of  motor,  204 

unit  of,  37 
Current  carrying  capacity  of  wire, 

543 
Curves,  directions  for  plotting,  529 

equations  for  straight  line,  533 
Cycle,  164 

definition  of,  430 

D 

J-connection,  513 
Damping     of     electrical     instru- 
ments, 202 

D' Arson val  galvanometer,  389 
Decade  bridge,  121 
Delta-connection,  513 
Depolarizer,  328 
Dial  bridge,  124 


Dielectric,  definition  of,  305 
power,  307 
strength  of,  306 
Difference  in  potential,  38 
Differential  motor,  221 
Direct  current,  hydraulic  analogy, 

39 

machines,  efficiency  of,  252 
elementary  form  of  genera- 
tor, 166 

resistance  of  armature,  176 
Direction,  of  E.M.F.  due  to  in- 
ductance, 275 
of  rotation  of  motor,  207 
Distribution  of  illumination  about 

a  lamp,  367 
Distributing      systems,      Edison 

three-wire,  241 
three-phase,  510 
two-phase,  505 
Drop  in  potential,  38 
Drum  armature,  169 
Dry  cell,  325 

E 
Eddy  current  loss  in  generator, 

189 

Edison  storage  battery,  354 
chemical  action  of,  355 
comparison    with     lead     type, 

357 

physical  changes  of,  356 
Edison  three-wire  system,  241 
Effective  value  of  A.C.  current,444 

of  E.M.F.,  445 

Effect  of  magnetic  field  on  mag- 
net, 12 
Efficiency,  86 

commercial,  257,  258 
depends  on  voltage  of  genera- 
tor, 232 
electrical,  257 
of  D.C.  machines,  252 
of  lamps,  88 
of  transmission,  87 
Efficiency  of  generators,  257 

of  motors,  258 
Electric  battery,  324 


550 


INDEX 


Electric  car  control,  220 

Electrical  charge,  303 

Electrical  efficiency  of  generators, 

257 

Electrical  equivalent  of  heat,  84 
Electricity,  negative,  303 

positive,  303 
Electrochemical  equivalent,  331 

table  of,  332 

Electro-dynamometers,  404 
Electrolysis,  331 

of  metal  water  mains,  etc.,  335 
Electrolyte,  325 
Electromagnetic    induction,    159, 

324 

Electromagnets,  31 
Electromotive  force  (E.M.F.),  50 
Electroplating,  333 
Electrotpying,  334 
E.M.F.,  direction  of  induced,  160 

effective  value  of  A.C.,  445 

induced,  269 
End  cell  control,  349 
Energy,  electrical  and  heat,  81 
Equation  for  straight  line,  533 
Equator  of  magnets.  2 
Equivalent  values,  table  of  elec- 
trical, mechanical,  and  heat 
units,  544 
Excitation  of  generator  field,  179 


Fall    of    potential    method     for 

measuring  resistance,  106 
Farad,  300 

Faure  type  storage  battery,  33S 
Feeders,  237 

Field  excitation  of  generator,  171) 
Field,  distribution  about  magnet,  4 

intensity,  3 

magnetic,  3 

Field  within  a  coil,  130 
Filament  of  incandescent     lamps, 

377 

Flaming  arcs,  374 
Flat-compounded  generator,  187 
Flux  (<£),  3 
Flux  density,  9,  131,  140 


Foot-candle,  365 
Foot-pound,  80 

Force  on  wire  in  magnetic  field,  199 
Forward  lead  of  brushes,  206 
Foucault  currents,  sec  Eddy  cur- 
rents 
Frequency,  definition  of,  430 


Galvanometer,  387 

astatic,  388 

Ayrton  universal  shunt,  393 

ballistic,  397 

constant,  118 

control,  389 

damping,  389,  390 

D' Arson val,  389 

dead  beat,  389 

sensibility  of,  395 

shunts,  391 

sine,  388 

working  constant  of,  395 
I    Gauss,  3 

I    General  law  for  A.C.  circuits,  473 
Generator,  field  excitation,  179 

A.C.  single  coil,  432 

effect  of  running  unlike  in  paral- 
lel, 249 

field,  3 

losses  in,  188 

resistance  of  D.C.  armature,  176 

types,  179 

voltage,   relation  to  efficiency 

of  transmission,  232 
Generator  effect  in  motors,  203 
Generator  voltage,  effect  on  size  of 

conductor,  235 
Gilbert,  153 

Gould  storage  battery,  339 
Gravity  control  of  solenoidal  gal- 
vanometer, 401 
Grid  of  storage  battery,  338 


H  (magnetizing  force),  131,  140 
H,  table  of  typical  values,  541 
Hand  rule,  for  generators,  160 
for  motors,  207 


INDEX 


551 


Heat  equivalent  of  electricity.  84 
Heat  units,   table  of  equivalent 

values,  544 
Henry,  289 
Hysteresis,  147 

constants,  152 

curve,  148 

loops   for   hard   and    annealed 
iron.  149 

loss,  computation  of,  150 

Steinmetz's  formula  for,  152 


PR,  76 
Illumination  363 

distribution  of   about  a    lamp, 
367 

intensity  of,  365 

measurement    of   intensity   of, 
370 

unit  of  intensity  of,  365 
Illuminometers,  370 
Impedance,  469 

of  parallel  combination,  486 
Incandescent  lamps,  377 

effect  of  voltage  variation  on 
candle-power   efficiency   and 
life  of,  37& 
Induced  E.M.F.,  269 

amount  of,  161 

direction  of,  160 

due  to  inductance  direction  of, 
275 

in  D.C.  armature;  178 
Inductance,  a  property  of  circuit, 
287 

computation  of,  462 
,  definition,  462 

of  transmission  lines,  294 
Inductance,  269 

mutual,  272 

Induction,  coils,  272,  276 
.    curve,  143 

definition,  7 

electromagnetic,  159 

mutual,  269 
Induction  motor,  497 
Inductivity,  307 


Inductive  circuits,  power  in,  494 
Inductive  reactance,  454 

computation  of,  462 
Insulation   resistance    of    motor, 
measured  by  voltmeter  meth- 
od, 116 
of  covered  wire,  measurement 

of,  117 

Intensity  of  light,  365 
Inter  pole  motor,  212 
Ironalosses  in  D.C.  machines,  253 


Joule,  80 
Jump  spark,  276 

K 

Kilowatt,  73 
Kilowatt-hour,  79 
Kirchhoff's  laws,  246 


Lagging  current,  450 
Lamination  of  cores,  189 
Lamps,  Cooper-Hewitt,  381 

mercury  arc,  381 

method  of  rating  incandescents, 
367 

Moore  tube,  383,  472 

Nernst,  380 

tantalum,  379 

tungsten,  379 

Law  of  inverse  squares,  366 
Lead  of  brushes?, ,  generator,  174 
Leading  current,  452 

cause  of,  459 
Lenz's  law,  270,  275 
Light,  intensity,  365 

nature  of,  364 

waves  of,  364 
Light  flux,  365 
Lines  of  force,  2 

nature,  property  of,  5 
Local  action  in  batteries,  329 
Locating  breaks  in  cable,  315 
Losses  in  D.C.  machines,  253  • 

in  generators,  188 
Luminous  arcs,  374 


552 


INDEX 


M 

,«,  140 

Magnetic  circuit,  2 

Magnetic  field,  24 

about  coil,  29 

about  straight  wire,  24 

resultant    of    circular    field    in 
parallel  field,  25 

rule  for  direction  of.  25 

within  a  coil,  130 
Magnetic  force  equations,  14 
Magnetic  hoist,  31 
Magnetism,  definiton  of,  1 

molecular  theory,  18 
Magnetization,  three  stages  of,  144 

curve,  143 

of  armature  of  generator,  170 
Magnetization    of    armature,    in 

motors,  205 
Magnetizing  force,  131,  140 

definition,  8 

Magnetomotive  force,  133 
Magnets,  ageing  of,  19 

definition  of,  1 

laws  of  attraction  and  repul- 
sion, 5 

permanent,  2,  17 

ring  magnets,  16 
Make  and  break  spark,  286 
Manganin,  use  in  ammeter  shunts, 

410 

Mean  horizontal  candle-power,  367 
Mean  spherical  candle-power,  307 
Measurement  of  resistance,  105 
Measuring  instruments,  387 
Mechanical  efficiency  of  motor,  258 
Mechanical    losses    in    D.C.    ma- 
chines, 253 

Mercury  arc  lamp,  381 
Mho,  57 
Microfarad,  301 
Mil,  96 
Mil-foot,  97 
Millivoltmeters,  408 

conversion  into  ammeters,  40S 

conversion  into  voltmeter,  411 
Moore  tube,  383 

illumination  of  silk,  472 


Motor  effect  in  generators,  201 

Motor  field,  31 

Moior  rheostat,  see  Starting  box 

Mutual  action  between  two  mag- 
nets, 13 

Mutual  inductance,  computation 
of,  293 

Mutual  induction,  269,  272 

Multipolar  generators,  175 

Murray  loop,  113 

N 
Negative  plate,  325 

of  battery  cells,  325 
Nernst  lamp,  380 
Neutral  axis,  172 

of  motor,  205 
No  field  release,  212 
Non-inductive  coils,  33 
No  voltage  release,  214 


Oersted,  135 
Ohm,  39 
Ohm's  law,  46 

applications,  48 

for  A.C.  circuits,  473 

in  A.C.  circuits,  448 

method  of  measuring  resistance, 
107 

of  magnetic  circuit,  135 
Open  circuit  cell,  329 
Ordinate,  definition  of,  529 
Over-compounded  generators,  187 
Overload  release,  215 


Parallel  A.C.  circuits,  current  and 

voltage  relations,  474 
Parallel  circuits,  52 

resistance,  voltage  and  current 

relation,  56 

Parallel    combinations    of    unlike 
generators     or     battery     cells, 

249 
Parallel  lighting  systems,  solution 

of,  59 


INDEX 


553 


Peltier  effects,  400 
Permeability,  140 

curve  of,  145 

definition  of,  6 

Permeability,  dependence  on  de- 
gree of  magnetization,  141, 
144 

vs.  reluctance,  137 
Phase  angle,  432 
Phase,  definition  of,  430 
Photometer,  368 

Bunsen.  368 

Sharp-Millar  universal,  371 
Photometry,  363 

units  of,  365 
Plante"    type  of    storage   battery, 

338  ' 

Polarization,  328 
Positive  plate,  325 

of  battery  cell,  area,  343 
Post  office  bridge,  119 
Potential,     fall    of     along    wire, 

45 

meaning  of,  42 
Potentiometer,  423 

calibration  of  voltmeter,  426 
Power,  73 

computation  of,  74 

in  A.C.  circuits,  492 

in  inductive  circuits,  494 

lost  in  setting  up    Eddy   cur- 
rents, 202 

measured  in  A.C.  circuits,  497 

measurement  of,  by  ammeters 
and  voltmeters,  413 

necessary  to  drive  generators, 

201 

Power  factor,  494 
Power  losses  in  generator,  188 
Power  units,  table  of  equivalent 

values,  544 
Primary  cell,  324 

chemical  action  of,  326 

tests  of,  330 
Pressure,  electric,  38 
Pulsating  current,  166 

curve  of,  167 
Pyrometer,  398 


R 

Reactance,  468 

capacity,  459 

capacity,  computation  of,  465 

inductive,  computation  of,  462 
Reactance,  inductive,  454 
Reaction  in  motor  armature,  205 
Refining  of  metals,  electrochem- 
ical, 334 

Regulating  coil  of  arc  lamps,  375 
Regulation,  definition,  210 
Relation  of  line  loss  to  generator 

voltage,  234 
Relation  of  size  of  conductor  to 

generator  voltage,  235 
Relation  of  voltage  of  generator  to 
efficiency  of  transmissiou,232 
Reluctance,  R,  computation  of,  137 

definition  of,  135 
Remanence,  148 
Resistance  of  wire,  table  of,  542 
Resistance,  39 

in  a  circuit,  95 

laws  of  parallel,  56 

laws  of  series,  56 

methods  of  measuring,  105 

of  D.C.  armature,  176 

of  wire,  97 
Resistivity,  of  copper,  98 

of  other  metals,  99 

table  of,  542 

Reversing  direction  of  motor,  207 
Ring  armature,  169 
Rotary  condenser,  515 
Ruhmkorff  coils,  277 

S 

Saturation,  of  iron,  142 
Saturation  point,  142 
Screens,  magnetic,  8 
Secondary  cells,  324 
Self-excited  generators,  179 
Self  inductance,  effect  of  in  A.C. 

circuits,  294 
Self  inductance,  285 

cause  of,  286 

computation  of,  288 

unit  of,  289 


554 


INDEX 


Self  induction,  282 

Separately     excited     generators, 

179,  180 
Series  A.C.  circuits,  current  and 

voltage  relations,  474 
Series  circuits,  52 

resistance,  voltage  and  current 

relation,  53 
Series  generator,  180 

external    characteristic    curve, 

182 
Series  motor,  216 

comparison  with  shunt,  217 
speed  (at  no  load),  216 
starting  directions,  219 
torque  of,  217 

torque  speed  characteristics,  218 
Series-parallel,  control  or,  electric 

cars,  220 

Shell-type  transformers,  278 
Shunt  generator,  183 

control  of  voltage  of,  185 
external    characteristic    curve, 

1.85 
Shunt    motor,    comparison    with 

series.  217 

direction  for  starting  and  stop- 
ping, 216 
torque,  217 
torque     speed     characteristics, 

218 
Siemen's       electro-dynamometer, 

404 

Significant  figures,  526 
Sine  curve  of  alternating  E.M.F., 

163,  435 

Sine,  defined,  539 
Sine  galvanometer,  388 
Size  of  wire,  table  of,  542 
Slide  wire  bridge,  111 

use    in    locating    cable    faults, 

112 

Sparking  of  motors,  206 
Sparkless  commutation,  of  genera- 
tor, 172 
of  motor,  206 

Specific  inductive  capacity,  307 
Specific  resistance,  98 


Speed  control,  of  shunt  motor,  210 
Speed  regulation,  of  shunt  motor, 

209 
"  Square  root  of  the  mean  squares ' ' 

value  of  A.C.  current,  444 
Starting-box,  need  of,  208 
Starting  resistance,  208 
Static  electricity,  303 
Step-up  transformers,  278 
Storage  batteries,  336 
construction,  338 
efficiency,  336 
types,  338 

Storage  batteries,  unlike,  in  paral- 
lel, 249 
Storage    battery,  advantage   and 

disadvantage  of,  348 
care  of  lead  cells,  342 
chemical     action     on     charge, 

341 
chemical    action   on  discharge, 

339 
comparison  of  lead  and  Edison 

cells,  357 

curves  of  exide  cell,  346,  347 
Edison  type,  354 
electrolyte  for,  344 
end-cell  control  of,  349 
floating  batteries,  352 
method  of  charging,  350 
normal  rate  of  charge  and  dis- 
charge, 343 
over-charge  and  over-discliarge, 

344 

rheostat  control  of,  349 
use  of  booster,  351 
use  on  constant  potential  line, 

348 

Storage  cell,  325 
Stow  device  for  control  of  motor 

speed,  211 

Straight  line,  equations  for,  533 
Stray  power.  252 

loss  in  shunt  generator,  254 
Strength  of  dielectric,  306 
Sucking  coils,  32 
Symbols,  of  electric  circuits,  39 
Synchronous  motor,  515 


INDEX 


555 


Tangent,  defined,  539 
Tantalum  lamps,  379 
Temperature  coefficient  of  resist- 
ance, table  of,  542 
Temperature  coefficient  of  resist- 
ance, 99,  542 

coefficient    for    various    initial 
temperatures,  101 

of  alloys,  104 

Temperature,    measured    by    re- 
sistance, 103 
Terminal  voltage,  50 
Thermo-bolometer,  398 
Thermo-couple,  398 
Thermo-electric  effects,  397 
Thermo-pile,  398 
Thomson  effects,  400 
Three-phase  distribution,  510 
Three-wire  system,  D.C.,  241 
Thumb  rule,  A.C.,  510,  518 

for  coil,  30 

for  wire,  25 
Torque,  199 

of  motors,  221 
Transformers,  278 
Trigonometric  functions    defined, 
538 

table  of  sin,  cos,  and  tan,  540 
Tungsten  lamps,  379 
Two-phase  distribution,  505 


U 


Unbalanced     three-wire    system, 

243 

Units,  table  of,  525 
Unit  field  strength,  11 
Unit  pole,  11 

relation  to  force  lines,  14 
Units,  commercial,  of  work,  79 
small,  of  work,  80 
standard  electrical,  61 
table  of  electrical,  mechanical, 

and  heat,  544 

Universal     galvanometer     shunt, 
393 


Useful  equations,  525 

Use  of  data  in  calculations,  527 


Variable  speed  motors,  211 
Varley  loop,  114 
Vector  diagram,  435 
Vector  sum,  475 
Volt,  38 

Voltage  of  generator  and  conse- 
quent line  loss,  234 
Voltage  regulation,    of   generator, 

184 

Voltameter,  335,  423 
Voltmeter,  411 

calibration  of  potential,  426 
connection  of,  71 
electrostatic,  412 
principles  of  operation,  Weston 

type,  27 
resistance  of,  72 
summary  of  types,  416 
Voltmeter-ammeter     method     of 

measuring  resistance,  107 
Voltmeter  method   of  measuring 
resistance,  116 


W 


Watt,  73 

Watt-hour  meter,  Thomson  inte- 
grating, 420 

Wattless  component,  496 
Wattmeter,  use  of,  78 

compensation  for  friction,  421 

compensation  in,  418 

creeping  of,  422 

electro-dynamometer,  417 

indicating,  417 

Thomson  integrating,  420 

Weston  type,  417 
Watt-second,  80 
Weston  D.C.  ammeters,  409 

A.C.  ammeter,  402 

A. C.  voltmeter,  403 

D.C.  voltmeter,  411 

wattmeter,  417 


INDEX 


Wheatstone  bridge,  107 

construction  of  various  types, 

119 

instruction  for  use  of,  110 
slide-wire  form,  111 
use  in  locating  cable  faults,  112 
Wheatstone    bridge    method     of 
measuring  resistance,  107 


Wire  tables  for  copper,  542 

use  of,  104 

Work,  commercial  units,  79 
Work  units,  table  of  equivalent 
values,  544 


Y-connection,  511 


DATE 


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WILL  BE  ASSESSED   FOR   FAILURE  TO   RETURN 
THIS    BOOK  ON   THE   DATE  DUE.   THE  PENALTY 

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